Answer
$\dfrac{y \sin x+2x}{\cos x-2y}$
Work Step by Step
Recall Equation 6 such as: $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$
Given: $y \cos x=x^2+y^2$
$\implies y \cos x-x^2-y^2=0$
Consider $F(x,y)=y \cos x-x^2-y^2=0$
$F_x=-y \sin x-2x\\F_y= \cos x-2y$
Equation (1) becomes:
$\dfrac{dy}{dx}=-\dfrac{(-y \sin x-2x)}{\cos x-2y}=\dfrac{y \sin x+2x}{\cos x-2y}$
