Answer
$\dfrac{e^y \cos x -1-y}{x-e^y \sin x}$
Work Step by Step
Recall Equation 6: $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$ ...(1)
Given: $e^y \sin x=x+xy$
$\implies e^y \sin x-x-xy=0$
Let us consider that $F(x,y)=e^y \sin x-x-xy=0$
$F_x=e^y \cos x -1-y\\F_y=e^y \sin x -x$
Equation (1) becomes:
$\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}=-\dfrac{e^y \cos x -1-y}{e^y \sin x -x}=\dfrac{e^y \cos x -1-y}{x-e^y \sin x}$
