Find the critical number of f on the given interval

Using the chain rule, $f'(x) = e^{-x^2}\cdot(-2x) = -2x\,e^{-x^2}$. Since $e^{-x^2} > 0$ for all $x$, the only way for $f'(x) = 0$ is if $-2x = 0$, giving $x = 0$. There are no points where $f'(x)$ is undefined, since $f'$ is a product of polynomial and exponential functions, both defined everywhere.

Therefore the only critical number of $f$ on $(-3,3)$ is $x = 0$.

For $x < 0$, $-2x > 0$ and $e^{-x^2} > 0$, so $f'(x) > 0$, meaning $f$ is increasing just to the left of $0$. For $x > 0$, $-2x < 0$, so $f'(x) < 0$, meaning $f$ is decreasing just to the right of $0$.

Since $f'$ changes from positive to negative at $x = 0$, the First Derivative Test tells us that $x = 0$ is a local maximum.

Evaluate $f$ at the critical number and both endpoints: $f(0) = e^{0} = 1$, $f(-3) = e^{-9}$, and $f(3) = e^{-9}$. Since $e^{-9}$ is an extremely small positive number, it is much less than $1$.

Comparing the three values, the largest is $f(0) = 1$. By the Closed Interval Method, the absolute maximum value of $f$ on $[-3,3]$ is $1$, attained at $x = 0$.