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