jmc

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 $x=\boxed{2}$.