<s>[INST] Proving that a function maintain a certain equation I need to prove that the function $$g(x,y,z)=f(\frac{1}{y}-\frac{1}{x},xye^{\frac{-z^2}{2}})$$ maintain the equation $$x^2g_x+y^2g_y=-\frac{x+y}{z}g_z$$ while $y\neq0$ and $f(u,v)$ differentiable for all $(u,v)$
How do i prove it without knowing what is $f$?
 [/INST] You start by writing $g_x$ as function of $f_u$ and $f_v$, thus
$$g_x = \frac{ \partial g}{\partial x}  = \frac{ \partial u}{\partial x} \frac{\partial f}{\partial u}  + \frac{ \partial v}{\partial x} \frac{\partial f}{\partial v} = \frac{ \partial \frac{1}{y} -\frac{1}{x}}{\partial x} f_u + \frac{ \partial xy \exp\left(- \frac{z^2}{2} \right)}{\partial x} f_v =x^{-2}f_u +y \exp\left(- \frac{z^2}{2} \right) f_v.$$
Likewise you find
$$g_y = - \frac{1}{y^2} f_u +x \exp\left(- \frac{z^2}{2} \right) f_v \textrm{ and } g_z = -zxy\exp\left(- \frac{z^2}{2} \right)f_v.$$
Then replace $g_x$, $g_y$ and $g_z$ in $$x^2 g_x + y^2 g_y + \frac{x+y}{z} g_z=0$$
by above formulas, and you see that the LHS is indeed $0$.
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