Answer
$\dfrac{dy}{dx}=\dfrac{ye^x -e^y}{xe^y -e^x}$
Work Step by Step
We have: $xe^y-ye^x=1$
We differentiate both sides with respect to $q$.
$xe^y \dfrac{dy}{dx}+e^y -ye^x -e^x \dfrac{dy}{dx}=0 \\ xe^y \dfrac{dy}{dx}-e^x \dfrac{dy}{dx}=y e^x -e^y \\ (xe^y -e^x) \dfrac{dy}{dx} =ye^x -e^y\\ \dfrac{dy}{dx}=\dfrac{ye^x -e^y}{xe^y -e^x}$
