Answer
$$-2re^{-r^{2}}e_{\mathbf{r}}$$
Work Step by Step
Given $$ f(x, y, z)=e^{-x^{2}-y^{2}-z^{2}}=e^{-r^{2}}$$
Since $$F(r)= e^{-r^2},\ \ \ \ \ \ \ \ F'(r) =-2re^{-r^{2}} $$
Then
\begin{align*}
\nabla f &=F^{\prime}(r) e_{\mathbf{r}}\\
&=-2re^{-r^{2}}e_{\mathbf{r}}
\end{align*}
