Consider the problems of maximizing u(x) subject to px = y and maximizing v(u(x)) subject to px = y, where v(u) is strictly increasing over the range of u. Prove that x* solves the first problem if and only if it also solves the second problem.

This is what I got, however i dont think its entirely correct.

To solve the first problem: maximizing u(x) subject to px = y
Assuming x is a one variable commodity bundle,
Lagrangean equation: L = u(x) - b(px-y)
First order condition:
dL/dx = U'x - bp = 0
dL/db = px - y =0
from the second equation, x* = y/p

To solve the second problem: maximizing v(u(x)) subject to px = y
Assuming x is a one variable commodity bundle,
Lagrangean equation: L = v(u(x)) - b(px-y)
First order condition:
dL/dx = V'*U'x - bp = 0
dL/db = px - y =0
from the second equation, x* = y/p

obviously, these two problems yield the same solution x* = y/p.
therefore, if x* solves the first problem if and only if it also solves the second problem.

Now, assuming x is a multi-variable commodity bundle,
To solve the second problem: maximizing u(x1, x2) subject to p1x1+p2x2 = y
Lagrangean equation: L = u(x1, x2) - b(px1+p2 -y)
First order condition:
dL/dX1 = U'x1 - bp1 = 0 (1)
dL/dX2 = U'x2 - bp2 = 0 (2)
dL/db = p1x1 +p2x2 - y =0 (3)
from the first two equations, U'x1 / U'x2 = p1/p2 (4)

Now, we turn to the second problem: maximizing v(u(x)) subject to px = y
Assuming x is a one variable commodity bundle,
Lagrangean equation: L = v(u(x)) - b(px-y)
First order condition:
dL/dX1 = V'*U'x1 - bp1 = 0 (5)
dL/dX2 = V'*U'x2 - bp2 = 0 (6)
dL/db = p1x1 +p2x2 - y =0 (7)
from the equations (5) and (6), U'x1 / U'x2 = p1/p2 (8)

then we find that equation system (7) and (8) are exactly the same as equation system (3) and (4), which means that these two problems yield the same solution set x*.
Therefore, if x* solves the first problem if and only if it also solves the second problem.