Answer
$f^{\prime}(x)=-2(2x^{3}+x)^{-3}(6x^{2}+1)$
Work Step by Step
Let $u(x)=x^{-2},\qquad v(x)=2x^{3}+x$
$\displaystyle \frac{du}{dx}=-2x^{-3}, \quad \frac{dv}{dx}=2(3x^{2})+1=6x^{2}+1$
Then,$\quad y=f(x)=u(v(x))$ and $f^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$
$\displaystyle \frac{du}{dv}=-2v^{-3}=-2(2x^{3}+x)^{-3}$
$f^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}=-2(2x^{3}+x)^{-2}\cdot(6x^{2}+1)$
