Answer
$(f^{-1})'(2)=\frac{1}{3}$
Work Step by Step
$f$ is one-to-one because $f'(x)=3x^2+3cosx-2sinx >0$ and so $f$ is increasing.
To use theorem 7, we need to know $f^{-1}(2)$ and we can find this by inspection:
$f(0)=2$ so $f^{-1}(2)=0$ thus by theorem 7, $(f^{-1})'(2)=\frac{1}{f'(f^{-1}(2))}=\frac{1}{f'(0)}=\frac{1}{3}$.