Answer
$$ z_x= \frac{-x}{\sqrt{9-x^2-y^2}}, \quad
z_y= \frac{-y}{\sqrt{9-x^2-y^2}}.$$
Work Step by Step
Recall that $(x^n)'=nx^{n-1}$
Since $ z=\sqrt{9-x^2-y^2} $, then by using the chain rule, we have
$$ z_x= \frac{-2x}{2\sqrt{9-x^2-y^2}}=\frac{-x}{\sqrt{9-x^2-y^2}}, \\
z_y= \frac{-2y}{2\sqrt{9-x^2-y^2}}=\frac{-y}{\sqrt{9-x^2-y^2}}.$$
