If $f^{-1}(g(x))=x^3-1$ and $g$ has an inverse, find $g^{-1}(f(7))$.
We know that $f^{-1}(u)=v$ is the same as $u=f(v)$.  Therefore $f^{-1}(g(x))=x^3-1$ is the same as  \[g(x)=f(x^3-1).\]We can also use that $g(s)=t$ is equivalent to $s=g^{-1}(t)$ to say \[x=g^{-1}(f(x^3-1)).\]This gives an expression containing $g^{-1}\circ f$.

Therefore, $g^{-1}(f(7))$ is the value of $x$ such that $x^3 - 1 = 7$.  Solving for $x$, we find $x = \boxed{2}$.