jmc

We are looking for some $x$ such that $f(x)=11$. We notice that by tripling $x$ we can increase $f(x)$ by 2 and also that $f(3)=5$.

Applying $f(3x)=f(x)+2$ repeatedly, we have: $$\begin{array}{rl}f(3)& =5\\ \Rightarrow f(9)& =7\\ \Rightarrow f(27)& =9\\ \Rightarrow f(81)& =11.\end{array}$$So $f^{-1}(11)=\boxed{81}$.