Answer
$\frac{dy}{dx}=\frac{ \sec{x^2y^2-2xy^2} }{2x^2y}$
Work Step by Step
$\sin{x^2y^2}=x$
$ \frac{d}{dx}(\sin{x^2y^2})=\frac{d}{dx}(x) $
$\cos({x^2y^2)} \frac{d}{dx}({x^2y^2})=1 $
$\cos({x^2y^2)} [y^2\frac{d}{dx}({x^2})+x^2\frac{d}{dx}y^2]=1 $
$\cos({x^2y^2)} [2xy^2+x^2\frac{d}{dy}y^2 \frac{dy}{dx}]=1 $
$\cos({x^2y^2)} [2xy^2+2x^2y \frac{dy}{dx}]=1 $
$ 2xy^2+2x^2y \frac{dy}{dx}=\frac{1}{ \cos({x^2y^2)} } $
$ 2xy^2+2x^2y \frac{dy}{dx}=\sec{x^2y^2}$
$ 2x^2y \frac{dy}{dx}=\sec{x^2y^2}-2xy^2$
$\frac{dy}{dx}=\frac{ \sec{x^2y^2-2xy^2} }{2x^2y}$
