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