Answer
$-\displaystyle \frac{3}{x^{4}}$
Work Step by Step
$\displaystyle \frac{d}{dx}[\frac{f(x)}{g(x)}]= \displaystyle \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$
$f(x)=\displaystyle \frac{1}{x}+\frac{1}{x^{2}}=x^{-1}+x^{-2}$
$f^{\prime}(x)=-x^{-2}-2x^{-3}=-\displaystyle \frac{1}{x^{2}}(1+\frac{2}{x})$
$=-\displaystyle \frac{1}{x^{2}}(\frac{2+x}{x})=-\frac{2+x}{x^{3}}$
$g(x)=x+x^{2}$
$g^{\prime}(x)=1+2x$
$f(x)==\displaystyle \frac{1}{x}+\frac{1}{x^{2}}=\frac{x+1}{x^{2}},$
$g(x)=x+x^{2}=x(1+x),$
$\displaystyle \frac{dy}{dx}=\frac{-\frac{2+x}{x^{3}}\cdot x(1+x)-(\frac{x+1}{x^{2}})(1+2x)}{x^{2}(1+x)^{2}}$
$=\displaystyle \frac{-\frac{(2+x)(1+x)}{x^{2}}-\frac{(x+1)(1+2x)}{x^{2}}}{x^{2}(1+x)^{2}}$
$=\displaystyle \frac{-(2+x)(1+x)-(x+1)(1+2x)}{x^{2}\cdot x^{2}(1+x)^{2}}$
$=\displaystyle \frac{(x+1)(-2-x-1-2x)}{x^{4}(1+x)^{2}}$
$=\displaystyle \frac{-3-3x}{x^{4}(1+x)}$
$=\displaystyle \frac{-3(1+x)}{x^{4}(1+x)}$
$=-\displaystyle \frac{3}{x^{4}}$
