Answer
$\displaystyle \frac{1}{9}$
Work Step by Step
Apply Th.1: derivative rule for inverses.
If $f(b)=a$, then
$\displaystyle \left.\frac{df^{-1}}{dx}\right|_{x=a}\ \ =\ \ \frac{1}{\left.\frac{df}{dx}\right|_{x=b}}$
We are given $f(3)=-1,\qquad (a=-1, b=3)$
$f(x)=x^{3}-3x^{2}-1$
$\displaystyle \frac{df}{dx}=3x^{2}-6x$
$\displaystyle \left.\frac{df^{-1}}{dx}\right|_{x=-1}\ \ =\ \ \frac{1}{\left.\frac{df}{dx}\right|_{x=3}}=\frac{1}{3(3^{2})-6(3)}=\frac{1}{27-18}=\frac{1}{9}$
