Answer
$\dfrac{-1 -y \cos (xy)}{1+x \cos (xy)}$
Work Step by Step
We have $x+y +\sin (xy)=1$
We differentiate both sides with respect to $x$.
$1+ \dfrac{dy}{dx}+\cos (xy) \dfrac{d}{dx} (xy)=0$
Simplify to obtain $\dfrac{dy}{dx}$ :
$1+\dfrac{dy}{dx}+x\cos (xy) \dfrac{dy}{dx} =-1 -y \cos (xy)$
or, $\dfrac{dy}{dx}=\dfrac{-1 -y \cos (xy)}{1+x \cos (xy)}$
