If \(y=1+x e^y\), then \(\frac{d y}{d x}=\)

\(\begin{aligned} & \mathrm{y}=1+\mathrm{xe}^{\mathrm{y}} \\ & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=0+\mathrm{xe}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{e}^{\mathrm{y}} \\ & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{xe}^{\mathrm{y}}-1\right)=-\mathrm{e}^{\mathrm{y}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{e}^{\mathrm{y}}}{\mathrm{xe}^{\mathrm{y}}-1} \\ & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{e}^{\mathrm{y}}}{\left(1+\mathrm{xe}^{\mathrm{y}}\right)-2}=\frac{-\mathrm{e}^{\mathrm{y}}}{\mathrm{y}-2}=\frac{\mathrm{e}^{\mathrm{y}}}{2-\mathrm{y}}\end{aligned}\)