jmc

Since $9$ can be written as $3^2$, we know that $3^{x^2+4x+4}=3^{2(x+2)}$ and $x^2+4x+4=2(x+2)$. Solving for $x$ we have: $$\begin{array}{r}{x}^2+4x+4=2x+4\\ \Rightarrow x^2+2x=0\\ \Rightarrow x(x+2)=0\end{array}$$So, $x=-2$ or $x=0$. Checking these solutions, we find that $3^0=9^0$ and $3^4=9^2$, which are both true statements. The sum of all possible values of $x$ is $-2+0=\boxed{-2}$.