Answer
$\dfrac{dy}{dx}=\dfrac{2xy+y^2}{-x^2-2yx}.$
Work Step by Step
$\dfrac{d}{dx}(x^2y)+\dfrac{d}{dx}(y^2x)=\dfrac{d}{dx}(-2)\rightarrow$
$(2xy+\dfrac{dy}{dx}(x^2))+(y^2+\dfrac{dy}{dx}(2yx))=0\rightarrow$
$2xy+y^2=\dfrac{dy}{dx}(-x^2-2yx)\rightarrow$
$\dfrac{dy}{dx}=\dfrac{2xy+y^2}{-x^2-2yx}.$
