Why don't you continue your previous similar question?
Anyway, this one is not well formulated. For x = 2, ==> f(2) +2* f(1/2) + 3*f(1) = 2, but f(1) is not defined!

@Dr Jean-Michel Collard  Yes, the functions u:=x->1/x; v:=x->x/(x-1);  generate o subgroup of order 6 (isomorphic to S_3) in the Moebius group. 

@maple fan 

1. For older versions, try (I don't have Maple 2015):

restart;
V:=<0,0,0>, <1,2,0>, <0,2,0>, <1,0,0>, <0,0,1>, <1,2,1>, <0,2,1>, <1,0,1>:
T:=<-1,0,0>, <0,-1,0>, <0,0,-1>: 
PM:=LinearAlgebra:-ProjectionMatrix([T[2]-T[1], T[3]-T[1]]):
Pr := v -> T[1]+PM.v:
Pr2:= v -> (PM.v)[[1,2]]:  # convexhull works in dimension 2.
P:=Pr~([V]):
P2:=convert~(Pr2~([V]),  list):
ch:=simplex:-convexhull(P2):
ind:=map(proc(u) local i; member(u, P2,'i'); i end,  ch):
C:=P[ind]: n:=nops(C):  # the vertices of the projection
Tri:=(v1,v2,v3) -> Surface( v1 + t*(v2+s*(v3-v2)-v1), s=0..1, t=0..1):
# the parametrization of the interior of a triangle
add(VectorCalculus:-SurfaceInt( x^2+y^2, [x,y,z] = Tri(C[1],C[i],C[i+1])), i=2..n-1);

2. The method works for polyhedra (the projections will be convex polygons). The projection of a cylinder is more complicated, with infinitely many extreme points in general.

3. A paremetrization of the triangle (v1,v2,v3) is   t*v1 + u*v2 + (1-t-u)*v3,  0<=t<=1, 0<=u<=1-t.
To have both parameters in [0,1], we may take  u = (1-t)*s.

@MathStudent0807 Otherwise, the function would not have existed.

@Nikol 

restart;
with(Statistics):
X := RandomVariable(Poisson(200.)):
F:=CDF(X, t):
a:=evalf[10](eval(F, t=50.)):
b:=evalf[20](eval(F, t=50.)):
a/b;
                        9.999999997 * 10^9 

@MathStudent0807 In the help page there is the surface integral of the function f(x,y,z) = y^2, over the sphere centered at <0,0,0> and having the radius r.

with(VectorCalculus):
SurfaceInt( y^2, [x,y,z] = Sphere( <0,0,0>, r ) );

Are you trying to say that your integral is much different?

Open the help page; type
?SurfaceInt

Note that you can open the help page as a worksheet, by clicking the 'WS' icon.
You will find there a very similar example; just edit it.

@tomleslie My remark is about the exact solution which is awful. It containe RootOfs for polynomials with degrees >30.

@acer You could also mention that for
expr := sin(x) - cos(32*x - 1/6*Pi);
solve(expr, allsolutions);
==> huge answer (length >1000000)
instead of a simple and easy to obtain:
{-1/93*Pi - 2/31*_Z1*Pi, 2/99*Pi - 2/33*_Z2*Pi}

For 
eq  := 2*exp(-exp(2*t)) + 4*t = 127

it will be (almost) impossible using fsolve to decide whether the solution is rational or not.

@rupsagar I don't see your desired result. Without it, it would be a waste of time to continue.

@rupsagar You forgot an example and the desired result.

@rupsagar Provide solution to what? If you use z(x) as above, the chain rule is automatic as shown.
If you want something else, provide a concrete example (include the desired result).

@mmcdara Yes, and all three of us know them. Anyway, the probability that someone knows the ode of the catenary but not its formula is low.

@mmcdara The "simpler" adjective is very relative:

restart;
Y := x -> (cosh(a*x)-1)/a:
L := x -> sinh(a*x)/a:
solve([L(x1)+L(x2)=140, Y(x1)=50, Y(x2)=70, a>0,x1>0,x2>0], explicit):
evalf(%);
#         {a = 0.09213375345, x1 = 26.14739561, x2 = 29.26983504}