Answer
$$\dfrac{e^y \cos x -1-y}{x-e^y \sin x}$$
Work Step by Step
We have: $F(x,y)=e^y \sin x-x-xy=0$
Re-arrange the equation as: $F(x,y)=e^y \sin x-x-xy=0$
Now , $$F_x=e^y \cos x -1-y \\ F_y=e^y \sin x -x$$
Since, $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$
Thus, we have:
$$\dfrac{dy}{dx}=-\dfrac{e^y \cos x -1-y}{e^y \sin x -x} \\=\dfrac{e^y \cos x -1-y}{x-e^y \sin x}$$
