Answer
$f^{\prime}(x)=-3(4x+1)(2x^{2}+x+1)^{-4}$
Work Step by Step
Let $u(x)=x^{-3},\qquad v(x)=2x^{2}+x+1$
$\displaystyle \frac{du}{dx}=-3x^{-4}, \quad \frac{dv}{dx}=2(2x)+1=4x+1$
Then,$\quad y=f(x)=u(v(x))$ and $f^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$
$\displaystyle \frac{du}{dv}=-3v^{-4}=-3(2x^{2}+x+1)^{-4}$
$f^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}=-3(2x^{2}+x+1)^{-4}\cdot(4x+1)$
