Answer
$\dfrac{\partial^2 z }{\partial x^2}=\dfrac{a^2}{y^2} \dfrac{\partial }{\partial y}(y^2\dfrac{\partial z}{\partial y})$
Work Step by Step
Let us suppose that $a=x-y; b=x+y$
Now, $z_x=\dfrac{1}{y}[a[f'(a)+ag'(b)] \implies z_{xx}=\dfrac{\partial^2 z}{\partial x^2}=\dfrac{a^2}{y}[f''(a)+g''(b)]$
and $z_{y}=-\dfrac{[f'(a)+f'(b)}{y^2}]+\dfrac{[-f(a)-f(b)}{y^2}]$
So, $[f''(a)+f''(b)]y ( \dfrac{a^2}{y^2} ) = [f''(a)+f''(b)] \times \dfrac{a^2}{y}$
So, it has been verified that
$\dfrac{\partial^2 z }{\partial x^2}=\dfrac{a^2}{y^2} \dfrac{\partial }{\partial y}(y^2\dfrac{\partial z}{\partial y})$
