Differentiation is not the whole story for this problem, though – there is a critical point at which the function is not differentiable.
A quick sketch of the curve should show that $\displaystyle x^2+4x<0$ for $\displaystyle -3<x<0$ and $\displaystyle x^2+4x>0$ for $\displaystyle 0<x<3$ and $\displaystyle f(x)$ has a nondifferentiable critical point at $\displaystyle x=0.$ Therefore a better solution is to rewrite the function as follows:
$\displaystyle f(x)\ =\ \begin{cases}-x^2-4x & -3<x\leqslant0\\
x^2+4x & 0\leqslant x<3\end{cases}$
You can then use differentiation on the open intervals $\displaystyle (-3,\,0)$ and $\displaystyle (0,\,3)$ separately to find the other critical point(s).