Answer
$ \dfrac{dy}{dx}=-\dfrac{y}{x+2y-xye^y -y^2 e^y}$
Work Step by Step
We have: $\ln (xy+y^2)=e^y$
We differentiate both sides with respect to $t$.
$\dfrac{1}{xy+y^2}(x\dfrac{dy}{dx}+y+2y \dfrac{dy}{dx})=e^y \dfrac{dy}{dx}\\ \dfrac{x}{xy+y^2}\dfrac{dy}{dx}+\dfrac{2y}{xy+y^2} \dfrac{dy}{dx})-e^y \dfrac{dy}{dx}=\dfrac{-y}{xy+y^2}\\(x+2y-xye^y-y^2e^y) \dfrac{dy}{dx}=-y$
Therefore, $ \dfrac{dy}{dx}=-\dfrac{y}{x+2y-xye^y -y^2 e^y}$
