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
Suppose $a=x-y; b=x+y$
Here, $\dfrac{\partial z}{\partial x}=\dfrac{1}{y}[a[f'(a)+ag'(b)]$
and $\dfrac{\partial^2 z}{\partial x^2}=\dfrac{a^2}{y}[f''(a)+g''(b)]$
$\dfrac{\partial z}{\partial y}=-\dfrac{[f'(a)+f'(b)}{y^2}]+\dfrac{[-f(a)-f(b)}{y^2}]$
and $[f''(a)+f''(b)]y \times \dfrac{a^2}{y^2}=\dfrac{a^2}{y} \times [f''(a)+f''(b)]$
Hence, it has been proved that
$\dfrac{\partial^2 z }{\partial x^2}=\dfrac{a^2}{y^2} \dfrac{\partial }{\partial y}(y^2\dfrac{\partial z}{\partial y})$
