jmc

If we substitute $f^{-1}(x)$ into our expression for $f$ we get $$f(f^{-1}(x))=4(f^{-1}(x){)}^3+1.$$This reduces to $$x=4(f^{-1}(x){)}^3+1.$$If we solve for $f^{-1}(x)$, we find that $f^{-1}(x)=\sqrt[3]{\frac{x-1}{4}}$. Therefore, $f^{-1}(33)=\sqrt[3]{\frac{33-1}{4}}=\sqrt[3]{8}=\boxed{2}$.