Answer
$$g'(x) = -(x-2)^{-2} .$$
Work Step by Step
Since $f(x)=2+x^{-1}$, then $f'(x)=-x^{-2}$. Hence by Theorem 2 and assuming that $g(x)=f^{-1}(x)$, we have
$$y=2+x^{-1}\Longrightarrow x=\frac{1}{y-2},$$
therefore,
$$g(x)=\frac{1}{x-2}, \quad g'(x)=\frac{1}{f'(g(x))}=-\frac{1}{ (1/(x-2))^{-2} }= -(x-2)^{-2} .$$
