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