Answer
$ \dfrac{dy}{dx}=-\dfrac{e^{-y}+y}{x-x e^{-y}-9}$
Work Step by Step
We have: $\dfrac{x}{e^y}+xy=9y$
We differentiate both sides with respect to $t$.
$-xe^{-y} \dfrac{dy}{dx}+e^{-y}+x \dfrac{dy}{dx} +y=9 \dfrac{dy}{dx}\\-xe^{-y} \dfrac{dy}{dx}+x \dfrac{dy}{dx} -9 \dfrac{dy}{dx}=-e^{-y}-y\\ (-x e^{-y} +x-9) \dfrac{dy}{dx} =-e^{-y}-y$
Therefore, $ \dfrac{dy}{dx}=-\dfrac{e^{-y}+y}{x-x e^{-y}-9}$
