Answer
\[y^{'}=-\frac{[y\cos (xy)+\sin (x+y) ]}{x\cos(xy)+\sin(x+y)}\]
Work Step by Step
$\sin (xy)=\cos (x+y)$ ___(1)
Differentiating (1) implicitly with respect to $x$
$\cos (xy)[x(y)^{'}+(x)^{'}y]=-\sin (x+y)[1+y^{'}]$
$xy^{'}\cos (xy)+y\cos (xy)=-\sin (x+y)-y^{'}\sin(x+y)$
$y^{'}[x\cos (xy)+\sin (x+y)]=-[\sin (x+y)+y\cos(xy)]$
\[y^{'}=-\frac{[\sin (x+y) +y\cos(xy) ]}{x\cos(xy)+\sin(x+y)}\]
\[y^{'}=-\frac{[y\cos (xy)+\sin (x+y) ]}{x\cos(xy)+\sin(x+y)}\]
Hence $y^{'}=-\Large\frac{[y\cos (xy)+\sin (x+y) ]}{x\cos(xy)+\sin(x+y)}$.
