jmc

We are looking for some $x$ such that $f(x)=64$. We notice that by doubling $x$ we can double $f(x)$ as well and also that $f(3)=1$.

Applying $f(2x)=2f(x)$ repeatedly, we have: $$\begin{array}{rl}f(3)& =1,\\ f(6)& =2,\\ f(12)& =4,\\ f(24)& =8,\\ f(48)& =16,\\ f(96)& =32,\\ f(192)& =64.\end{array}$$So $f^{-1}(64)=\boxed{192}$.