Can anyone solve this problem based on Euler's theorem on homogenous functions

Prove that : x²∂²u/∂x² + 2xy ∂²u/∂x∂y + y² ∂²u/∂y² = g(u) [g'(u) - 1]

where g(u) = n f(u)/f'(u)

By Euler's formula: x ∂u/∂x + y ∂u/∂y = n. f(u)/f'(u)
its partial differentian, I need it urgent.

given that x ∂u/∂x + y ∂u/∂y = n. f(u)/f'(u) and g(u) = n f(u)/f'(u)

hence x ∂u/∂x + y ∂u/∂y = g(u).............(1), (pretty obvious)

now differentiating (1) partially w.r.t 'x',

[ x ∂²u/∂x² + ∂u/∂x . 1] + y ∂²u/∂x∂y = g'(u) ∂u/∂x

=> x ∂²u/∂x² + y. ∂²u/∂x∂y = [g'(u) - 1] ∂u/∂x....................(2)

similarly, on differentiating (1) partially w.r.t 'y'.

y ∂²u/∂y² + x. ∂²u/∂y∂x = [g'(u) - 1] ∂u/∂y.........................(3)

Multiplying (2) by x and (3) by y and adding,

x² ∂²u/∂x² + 2xy ∂²u/∂x∂y + y² ∂²u/∂y² = [g(u) - 1] [ x ∂u/∂x + y ∂u/∂y]
= [g'(u) - 1] g(u) from (1)
= g(u) [g'(u) - 1] Proved.

u = x^a*y^b

x²∂²u/∂x² = a(a-1)u
2xy ∂²u/∂x∂y = ab u
y²∂²u/∂y² = b(b-1)u

==>

[a(a-1) + ab + b(b-1)]u = g(u) [g'(u) - 1]

A solution to this is given by g(u) = c*u, where c satisfies


[a(a-1) + ab + b(b-1)] = c(c-1)

Also, since g = c*u, we see that f satisfies

g = n*f/f' ==>
f' = n/(cu) * f

which has a solution given by

f = A*u^(n/c).

and so, f(x,y) = A* x^(n*a/c) * y^(n*b/c).

Without more information, a, b and c cannot be determined.