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