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