Answer
$\frac{dy}{dx} = \frac{\sin y^2 - 2yx\cos x^2}{\sin x^2-2xy\cos y^2}$
Work Step by Step
$y\sin x^2 = x \sin y^2$
Differentiate both sides with respect to x
$\frac{dy}{dx}\sin x^2 + 2yx\cos x^2 =\sin y^2 + 2xy\cos y^2 \frac{dy}{dx} $
Isolate $\frac{dy}{dx}$
$\frac{dy}{dx}\sin x^2 - 2xy\cos y^2 \frac{dy}{dx} = \sin y^2 - 2yx\cos x^2$
$\frac{dy}{dx}(\sin x^2-2xy\cos y^2) = \sin y^2 - 2yx\cos x^2$
$\frac{dy}{dx} = \frac{\sin y^2 - 2yx\cos x^2}{\sin x^2-2xy\cos y^2}$