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