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