Answer
$y^{\prime}=-\frac{x}{1+x}$
Work Step by Step
$y=\ln \left(e^{-x}+x e^{-x}\right)\\
y^{\prime}=\frac{1}{e^{-x}+x e^{-x}} \times\left(-\frac{1}{e^{x}}+(x)\left(-e^{-x}\right)+\left(e^{-x}\right)(1))\right.\\
y^{\prime}=\frac{1}{e^{-x}+x e^{-x}} \times\left(\frac{-1-x+1}{e^{x}}\right)\\
y^{\prime}=-\frac{x}{1+x}$
