I'm trying show that
$$\frac{\partial^2z}{\partial x^2}=\frac{a^2}{y^2}\frac{\partial}{\partial y}( y^2\frac{\partial z}{\partial y})$$knowing that: $$ z=\frac{1}{y}(f(ax+y)+g(ax-y)). $$ I know that, to do this, I need to use partial differentiation, but I just have no idea where to start. Is there any way I can simplify $f(ax+y)$ and $g(ax-y)$?
This is just differentiating.
Since $z=\frac1y[f(ax+y)+g(ax-y)]$, we clearly have $$ \frac{\partial^2 z}{\partial x^2}=\frac1y[a^2f''(ax+y)+a^2g''(ax-y)] $$ on the LHS. For the RHS, we start by differentiating $z$ with respect to $y$: $$ \frac{\partial z}{\partial y}=-\frac1{y^2}[f(ax+y)+g(ax-y)]+\frac1y[f'(ax+y)-g'(ax-y)] $$ So $$ y^2\frac{\partial z}{\partial y}=-[f(ax+y)+g(ax-y)]+y[f'(ax+y)-g'(ax-y)] $$ Differentiate with respect to $y$ again, \begin{align*} \frac{\partial}{\partial y}\left(y^2\frac{\partial z}{\partial y}\right)&=-[f'(ax+y)-g'(ax-y)]\\ &\quad+[f'(ax+y)-g'(ax-y)]+y[f''(ax+y)+g''(ax-y)]\\ &=y[f''(ax+y)+g''(ax-y)] \end{align*} Hence the RHS is $$ \frac{a^2}{y^2}y[f''(ax+y)+g''(ax-y)]=\frac{a^2}{y}[f''(ax+y)+g''(ax-y)] $$ So indeed $LHS=RHS$.