Answer
$\dfrac{e^y \cos x -1-y}{x-e^y \sin x}$
Work Step by Step
We are given that $e^y \sin x=x+xy$
Re-arrange as: $e^y \sin x-x-xy=0$
Consider $F(x,y)=e^y \sin x-x-xy=0$
$F_x=e^y \cos x -1-y$ and $F_y=e^y \sin x -x$
Use Equation 6 which is: $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$
This implies that
$\dfrac{dy}{dx}=-\dfrac{e^y \cos x -1-y}{e^y \sin x -x}=\dfrac{e^y \cos x -1-y}{x-e^y \sin x}$
