If $cos^2y+sin\, xy =\lambda, $ then $\frac{dy}{dx}=$

$\frac{ycosxy}{sin2y-xcosxy}$

Given: $\cos^2 y + \sin(xy) = \lambda$

Differentiating both sides with respect to $x$:

$\frac{d}{dx}(\cos^2 y) + \frac{d}{dx}(\sin(xy)) = \frac{d}{dx}(\lambda)$

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

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

Collecting $\frac{dy}{dx}$ terms:

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

Solving for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-y \cos(xy)}{-2 \cos y \sin y + x \cos(xy)}$

$\frac{dy}{dx} = \frac{y \cos(xy)}{2 \cos y \sin y - x \cos(xy)}$