Answer
$f^{\prime}(x)=-(x-1)^{-2}$
Work Step by Step
Let $u(x)=x^{-1},\qquad v(x)=x-1$
$\displaystyle \frac{du}{dx}=-x^{-2}, \quad \frac{dv}{dx}=1$
Then,$\quad y=f(x)=u(v(x)$ and
$f^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$
$\displaystyle \frac{du}{dv}=-v^{-2}=-(x-1)^{-2}$
$f^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}=-(x-1)^{-2}\cdot 1=-(x-1)^{-2}$