Answer
$$\frac{\partial^2 f}{\partial x\partial y}=(x-y)^{-1}+y(x-y)^{-2}-x(x-y)^{-2}-2xy(x-y)^{-3}.$$
Work Step by Step
Since $ g(x,y)=\frac{xy}{x-y}=xy(x-y)^{-1}$, then by using the product rule, we have
$$\frac{\partial f}{\partial x}= y(x-y)^{-1}-xy(x-y)^{-2}.$$
Hence, again by using the product rule, we get:
$$\frac{\partial^2 f}{\partial x\partial y}=(x-y)^{-1}+y(x-y)^{-2}-x(x-y)^{-2}-2xy(x-y)^{-3}.$$
