Answer
$$ z_x= \frac{-y}{(x-y)^2}, \quad
z_y= \frac{x}{(x-y)^2}.$$
Work Step by Step
Recall that $(x^n)'=nx^{n-1}$
Since $ z=\frac{x}{x-y} $, then by using the quotient rule, we have
$$ z_x= \frac{(x-y)(1)-x(1)}{(x-y)^2}=\frac{-y}{(x-y)^2}, \\
z_y= \frac{(x-y)(0)-x(-1)}{(x-y)^2}=\frac{x}{(x-y)^2}.$$
