Answer
$$\dfrac{y \sin x+2x}{\cos x-2y}$$
Work Step by Step
We have: $y \cos x=x^2+y^2$ and $F=y \cos x-x^2-y^2=0$
So, $$F_x=-y \sin x-2x \\ F_y= \cos x-2y$$
Since, $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$
Thus,
$$\dfrac{dy}{dx}=-\dfrac{(-y \sin x-2x)}{\cos x-2y} \\=\dfrac{y \sin x+2x}{\cos x-2y}$$
