Answer
$\frac{dy}{dx}=\frac{xy^2-x}{y-yx^2}$
Work Step by Step
$x^2y^2=x^2+y^2\\ \frac{dy}{dx}(x^2y^2=x^2+y^2)\\ 2xy^2+2yx^2\frac{dy}{dx}=2x+2y\frac{dy}{dx}\\ 2xy^2-2x=2y\frac{dy}{dx}-2yx^2\frac{dy}{dx}\\ 2xy^2-2x=\frac{dy}{dx}(2y-2yx^2)\\ \frac{dy}{dx}=\frac{2xy^2-2x}{2y-2yx^2}\\ \frac{dy}{dx}=\frac{2(xy^2-x)}{2(y-yx^2)}\\ \frac{dy}{dx}=\frac{xy^2-x}{y-yx^2}$