jmc

Substituting $f^{-1}(x)$ into our expression for $f$, we get $$f(f^{-1}(x))=\frac{16}{5+3f^{-1}(x)}.$$Since $f(f^{-1}(x))=x$ for all $x$ in the domain of $f^{-1}$, we have $$x=\frac{16}{5+3f^{-1}(x)}.$$When $x=2$ this says $$2=\frac{16}{5+3f^{-1}(2)}.$$Solving for $f^{-1}(2)$, we find $f^{-1}(2)=1$. Then $[f^{-1}(2){]}^{-2}=1^{-2}=\boxed{1}$.