Answer
$ \dfrac{dy}{dx}=\dfrac{y-y^2}{3y-1-y^2}$
Work Step by Step
We have: $\ln (y^2-y) +x=y$
We differentiate both sides with respect to $t$.
$\dfrac{1}{y^2-y}(2 \dfrac{dy}{dx}- \dfrac{dy}{dx}) +1= \dfrac{dy}{dx}\\\dfrac{2y}{y^2-y} \dfrac{dy}{dx}-\dfrac{1}{y^2-y} \dfrac{dy}{dx} +1= \dfrac{dy}{dx} \\ \dfrac{2y}{y^2-y} \dfrac{dy}{dx}-\dfrac{1}{y^2-y} \dfrac{dy}{dx} - \dfrac{dy}{dx}=-1\\ (\dfrac{2y}{y^2-y} -\dfrac{1}{y^2-y} -1 ) \dfrac{dy}{dx}=-1 $
Therefore, $ \dfrac{dy}{dx}=\dfrac{y-y^2}{3y-1-y^2}$
