Answer
$$
z_x= y^x\ln y,\quad
z_y= \frac{x}{y}y^x.
$$
Work Step by Step
Recall the product rule: $(uv)'=u'v+uv'$
Recall that $(\ln x)'=\dfrac{1}{x}$
Since $ z=y^x $, applying $\ln $ on both sides, we get
$$\ln z=x\ln y.$$
Now, by using the product rule, we have
$$
\frac{z_x}{z}=\ln y \Longrightarrow z_x=z\ln y=y^x\ln y,\\
\frac{z_y}{z}=\frac{x}{y} \Longrightarrow z_y=z \frac{x}{y}=\frac{x}{y}y^x.
$$
