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