jmc

Substituting $f^{-1}(x)$ into our expression for $f$, we find $$f(f^{-1}(x))=\frac{3f^{-1}(x)+2}{5}.$$Since $f(f^{-1}(x))=x$ for all $x$ in the domain of $f^{-1}$, we have $$x=\frac{3f^{-1}(x)+2}{5}.$$Solving for $f^{-1}(x)$, we obtain $f^{-1}(x)=\frac{5x-2}{3}$. In particular, $f^{-1}(4)=\frac{5\cdot 4-2}{3}=18/3=6$, so $[f^{-1}(4){]}^{-1}=\boxed{\frac{1}{6}}$.