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
Let us suppose that $a=x-y$ and $ b=x+y$
Now, $\dfrac{\partial z}{\partial x}=-\dfrac{[f(a)+f(b)}{x^2}+\dfrac{[f'(a)+f'(b)}{x}$
$\implies x^2[\dfrac{\partial z}{\partial x}]=-x^2[\dfrac{[f(a)+f(b)}{x^2}+\dfrac{[f'(a)+f'(b)}{x}]$
Next, $z_{xx}=x[f''(a)+g''(b)]$ and $z_y=x[-\dfrac{[f'(a)+f'(b)}{x^2}]$
and $z_{yy}=\dfrac{x[f''(a)+g''(b)]}{x^2} \times x^2$
So, it has been verified that
$\dfrac{\partial }{\partial x}(x^2 \dfrac{\partial z}{\partial x})=x^2 \dfrac{\partial^2 z}{\partial y^2}$
