${\sin ^2}y + \cos xy = k$

We are given that, ${\sin ^2}y + \cos xy = k$
Differentiating (i) on both sides w.r.t. x, we get

$2\sin y\cfrac{d}{{dx}}(\sin y) + ( - \sin xy)\cfrac{d}{{dx}}(xy) = 0$ 2

$\Rightarrow$ $2\sin y\cos y\cfrac{{dy}}{{dx}} + ( - \sin xy)\left[ {x\cfrac{{dy}}{{dx}} + y} \right] = 0$

$\Rightarrow$ $2\sin y\cos y\cfrac{{dy}}{{dx}} - x\sin xy\cfrac{{dy}}{{dx}} - y\sin xy = 0$

$\Rightarrow$ $\cfrac{{dy}}{{dx}}[2\sin y\cos y - x\sin xy] = y\sin xy$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = \cfrac{{y\sin xy}}{{\sin 2y - x\sin xy}}$