Answer
\begin{aligned} \frac{\partial^{2} z}{\partial x^{2}} &= -\frac{1}{(x-y)^{2}} \\ \frac{\partial^{2} z}{\partial y \partial x} &=\frac{1}{(x-y)^{2}} \end{aligned}
$$\frac{\partial^{2} z}{\partial x \partial y}=\frac{\partial^{2} z}{\partial y \partial x}=\frac{1}{(x-y)^{2}}$$
Work Step by Step
Given $$z =\ln (x-y)$$
So, we get
\begin{aligned} \frac{\partial z}{\partial x} &=\frac{1}{x-y}=(x-y)^{-1 }\\ \frac{\partial^{2} z}{\partial x^{2}} &=(-1)(x-y)^{-1-1 }=-\frac{1}{(x-y)^{2}} \\ \frac{\partial^{2} z}{\partial y \partial x} &=(-1)(-1)(x-y)^{-1-1 }=\frac{1}{(x-y)^{2}} \end{aligned}
and we have
\begin{aligned} \frac{\partial z}{\partial y} &=\frac{(-1)}{x-y}=\frac{1}{y-x} =(y-x)^{-1}\\ \frac{\partial^{2} z}{\partial y^{2}} &=(-1)(y-x)^{-1-1}=-\frac{1}{(x-y)^{2}} \\ \frac{\partial^{2} z}{\partial x \partial y} &=(1)(y-x)^{-1-1}=\frac{1}{(x-y)^{2}} \end{aligned}
we can observe that
$$\frac{\partial^{2} z}{\partial x \partial y}=\frac{\partial^{2} z}{\partial y \partial x}$$
