Answer
$f^{\prime}(x)=-4x(2x^{2}-2)^{-2}$
Work Step by Step
Let $u(x)=x^{-1},\qquad v(x)=2x^{2}-2$
$\displaystyle \frac{du}{dx}=-x^{-2}, \quad \frac{dv}{dx}=2(2x)=4x$
Then,$\quad y=f(x)=u(v(x))$ and $f^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$
$\displaystyle \frac{du}{dv}=-v^{-2}=-(2x^{2}-2)^{-2}$
$f^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}=-(2x^{2}-2)^{-2}\cdot(4x)$