The sphere can be tangent to (some of) the edges in the exterior of the tetrahedron. In this case, Kitonum's conditions must be adjusted.

with(plots):with(plottools):
display(tetrahedron([[0.,0.,0.],[3.,0.,0.],[-4.,3.,0.],[-2.502782406,1.,4.628518959]],transparency=0.2),
        sphere([7.807886553,23.42365966,8.196378356],24.81629404,transparency=0.8,color=yellow),axes=none,
        orientation=[60,70,150]);

not(symbol) in this context is the type of an expression obtained by applying the function not to a symbol.
Examples of this type are not(x), not(Pi).

The type  "not a symbol"  would be Not(symbol).
type("abc", Not(symbol)); # true
type("abc", not(symbol)); # false
 

restart;
H:= k -> < cos(2*k/n*Pi), sin(2*k/n*Pi) >:
n:=100; dt:=0.005; t:=0;
P:=proc(k) global t; Threads:-Sleep(t);t:=t+dt; 
  plots:-display(plot([cos,sin,0..2*Pi]), plots:-arrow(<0,0>,H(k))) end:
Explore(P(k), k=1..n, animate, numframes=n);

Q:=proc(k) 
  plots:-display(plot([cos,sin,0..2*Pi]), plots:-arrow(<0,0>,H(k))) end:
Explore(Q(sqrt(k)), k=1..n^2, animate, numframes=5*n);

This is straightforward (unlike your previous question).

asympt(int((1-t^2)^n, t=0..1),n);

    sqrt(Pi)*sqrt(1/n)/2 + ...

It works for me.

restart;
F:=[-9*a[8]*a[7]*a[9]-5/2*a[3]*a[4]*a[5]*a[9]+6*a[8]^3+5/2*a[3]*a[5]^2*a[8]+3*a[9]
^2*a[6]+3*a[4]^3*a[9]-9*a[4]*a[9]-4*a[4]^2*a[5]*a[8]+3/2*a[4]*a[7]*a[5]^2+9*a[5
]*a[8]-1/2*a[5]^3*a[6], 2*a[3]^4*a[5]-2*a[3]^3*a[4]^2+24*a[3]^2*a[7]^2-48*a[3]*
a[4]*a[6]*a[7]-12*a[3]*a[5]*a[6]^2+36*a[4]^2*a[6]^2+9*a[3]^3-54*a[6]^2, 11*a[5]
^2*a[6]*a[8]+20/3*a[3]*a[5]^2*a[4]^2-12*a[9]*a[4]^2*a[7]-35*a[5]*a[8]^2*a[3]+24
*a[4]^2*a[5]-13/3*a[4]^4*a[5]-90*a[8]^2-11*a[6]*a[4]*a[5]*a[9]+5*a[4]*a[5]*a[7]
*a[8]+10*a[3]*a[7]*a[9]*a[5]+10*a[3]*a[8]*a[9]*a[4]-7/3*a[5]^3*a[3]^2+22*a[8]^2
*a[4]^2+3/2*a[9]^2*a[3]^2-15*a[3]*a[5]^2-3/2*a[7]^2*a[5]^2+36*a[9]*a[7]-27*a[5]
, 10*a[5]*a[3]*a[6]*a[8]+10*a[4]*a[6]*a[7]*a[5]-90*a[7]^2-12*a[6]*a[4]^2*a[8]+
20/3*a[5]*a[3]^2*a[4]^2+11*a[3]^2*a[7]*a[9]-35*a[3]*a[7]^2*a[5]+22*a[7]^2*a[4]^
2+3/2*a[6]^2*a[5]^2-7/3*a[3]^3*a[5]^2-3/2*a[3]^2*a[8]^2+36*a[6]*a[8]+24*a[4]^2*
a[3]-13/3*a[4]^4*a[3]-15*a[5]*a[3]^2+5*a[7]*a[3]*a[4]*a[8]-11*a[3]*a[4]*a[6]*a[
9]-27*a[3], 24*a[3]*a[4]*a[7]*a[9]-87/2*a[3]*a[7]*a[5]*a[8]+24*a[6]*a[4]*a[5]*a
[8]+33/2*a[3]*a[6]*a[5]*a[9]+54*a[4]^3-9*a[4]^5+9/2*a[8]*a[3]^2*a[9]+51*a[4]^2*
a[7]*a[8]+9/2*a[7]*a[5]^2*a[6]-7*a[3]^2*a[5]^2*a[4]+16*a[4]^3*a[5]*a[3]-45*a[6]
*a[4]^2*a[9]-18*a[3]*a[4]*a[8]^2-45*a[5]*a[3]*a[4]-18*a[4]*a[7]^2*a[5]-81*a[4]+
81*a[6]*a[9]-135*a[7]*a[8], 138*a[6]*a[7]*a[4]*a[9]-129*a[3]*a[4]^2*a[7]*a[5]-\
147*a[6]*a[7]*a[5]*a[8]+187/2*a[3]^2*a[4]*a[5]*a[8]-360*a[4]^2*a[7]+62*a[7]*a[4
]^4+27*a[3]^2*a[9]+18*a[7]^3*a[5]-48*a[3]*a[6]*a[8]*a[9]-129/2*a[3]*a[6]*a[5]^2
*a[4]-243*a[6]*a[4]*a[5]+85*a[4]^3*a[5]*a[6]-21*a[3]^2*a[4]^2*a[9]+288*a[8]*a[3
]*a[4]+288*a[5]*a[3]*a[7]+91/2*a[7]*a[5]^2*a[3]^2-76*a[4]^3*a[8]*a[3]-24*a[3]*a
[7]^2*a[9]+192*a[3]*a[7]*a[8]^2-132*a[6]*a[4]*a[8]^2+33*a[6]^2*a[5]*a[9]+9/2*a[
3]^3*a[5]*a[9]-30*a[7]^2*a[4]*a[8]+486*a[7], 270*a[6]^2*a[4]*a[9]-3*a[3]^2*a[4]
^2*a[8]-33/2*a[3]^3*a[4]*a[9]+57/2*a[3]^3*a[5]*a[8]+261*a[3]*a[7]^2*a[8]-72*a[6
]^2*a[5]*a[8]+621*a[7]*a[3]*a[4]+54*a[3]*a[6]*a[8]^2+405*a[5]*a[3]*a[6]+87/2*a[
3]^2*a[5]^2*a[6]-219*a[4]^3*a[7]*a[3]-1134*a[6]*a[4]^2+270*a[6]*a[4]^4+108*a[3]
^2*a[8]+108*a[4]*a[7]^3+363/2*a[4]*a[3]^2*a[7]*a[5]-171*a[3]*a[6]*a[7]*a[9]-450
*a[6]*a[4]*a[7]*a[8]-285*a[6]*a[4]^2*a[5]*a[3]+972*a[6]]:
A:=indets(F)[];
X:=seq(x||i,i=3..9);
G:=Groebner:-Basis(F,tdeg(A));
FX:=eval(F, [A=~X]):
GX:=Groebner:-Basis(FX,tdeg(X));
evalb( eval(G, [A=~X]) = GX);  # true

BTW, I don't recommend using the document mode (and 2D input) for programming. For example it is hard to run your worksheet line by line.

As mmcdara said, this can be computed using the Laplace method. Unfortunaly he made a mistake, the maximum point found was outside the interval of integration.

restart;
J:=  Int(exp(-n*(x*cosh(t)+t)), t = 0 .. infinity):
h:= convert(series(-n*(x*cosh(t)+t),t,3), polynom):
int(exp(h),t=0..infinity) assuming x>0,n>0:
A:=asympt(%,n,2) assuming x>0;

     A := (1/n+O(1/n^2))/exp(n*x)

evalf( eval([J=A], [x=1, n=100]));  # check
evalf( eval([J=A], [x=1/4, n=100]));

     [3.683935796*10^(-46) = 3.720075976*10^(-46)+3.720075976*10^(-44)*O(1/10000)]
     [1.385347786*10^(-13) = 1.388794386*10^(-13)+1.388794386*10^(-11)*O(1/10000)]

For simplicity, I will suppose that f : (0, oo) --> R  is continuous and the relation

Int(f(x), x=a..b) = Int(f(x), x=1/b..1/a)    (*)

holds for all 0 < a < b < oo.

(*) is equivalent to 

Int(f(x) - f(1/x)/x^2 , x=a..b) = 0, for all a,b>0
so, f(x) - f(1/x)/x^2 = 0, for all x>0.

The general form for f is given by:

f(x) = piecewise(x<1, u(x), u(1/x)/x^2),

where u : (0,1] --> R is an arbitrary continuous function.

Note. Your example f(x) = 1/(x^2+1) satisfies f(1/x^2)/x^2 = f(x).

You could use an external program, able to extract text information from windows (and their children). and/or simulate mouse clicks. E.g. AutoIt. I have used long time ago for similar purposes Visual DialogScript.

The exact and numeric solutions agree.
Actually, 
solve(helpode_solution2,y(x));

==> y(x) = sqrt(x^2+1);
There is also  - sqrt(x^2+1), but we want y(0)=1.

Note also that if you use

helpode_solution := combine(simplify(convert(helpode_solution,ln))) assuming x>0 

then the division by 0 disappears.

a:=fsolve(1/FINT2);
#                       a := 0.02355400734
e:=1e-6:
int(FINT2, x = 0 .. a-e, numeric);
#                         -0.7603068518
int(FINT2, x = a+e .. 1, numeric);
#                          1.155767949
%+%%;
#                          0.3954610972

You may use the operator D; just fill its remember table

D(y):=1: D(x):=x*y: D(z):=z^2:
D(x*y + y*z);
D(1/2*x*y^2 + 3*y^3*z + a); # you may want D(a):=0;

y(x) = 0 is obtained for _C1 --> infinity.
This is documented (?dsolve,details).

What you write here is simply wrong.

ode:=diff(y(x), x)/y(x);
sol:=y(x)=0;

algsubs(sol,ode);   #this gives ZERO. It should be 1

algsubs should return an error (division by 0). 1 is out of the question.

ode:=diff(y(x),x)^2+2*x*diff(y(x),x)/y(x)-1 = 0;
sol:=y(x)=0;
odetest(sol,ode);
         0

is also wrong (division by 0 too). y(x)=0 is not a solution (odetest checks the solution for a transformed ode, which is not equivalent with the original).

You say: Which is 0/0 but this is 1 in the limit. This is nonsense.

You have changed the problem, now you want  f(t*x,t^p*y) = t^(p-1)*f(x,y).
In this case the procedure is even simpler:

Q := (F,x,y) -> simplify( (F + diff(F,x)*x)/(F - diff(F,y)*y) ):

Let's compare with the previous P:

P := (F,x,y)-> simplify(((-x*diff(F,x,x)-diff(F,x))*F+diff(F,x)^2*x)/(y*(F*diff(F,x,y)-(diff(F,x))*(diff(F,y))))):

Q(-2*x/3+sqrt(x^2+3*y)/3, x,y ),   P(-2*x/3+sqrt(x^2+3*y)/3, x,y );
#                              2, 2

Q(3*sqrt(x*y), x,y),   P(3*sqrt(x*y), x,y);
#                             3, -1

P := (F,x,y)-> simplify(((-x*diff(F,x,x)-diff(F,x))*F+diff(F,x)^2*x)/(y*(F*diff(F,x,y)-(diff(F,x))*(diff(F,y))))):
P(-2*x*(1/3)+(1/3)*sqrt(x^2+3*y), x,y);
#                              2
P(x*y^3-x^2/y^3, x,y);
#                               1/6