could you tell me how to attack this one
$\displaystyle \frac{d}{dx} (x^2y^2+2y=x^3)$
$\displaystyle x^2 \cdot 2y \cdot \frac{dy}{dx} + y^2 \cdot 2x + 2 \cdot \frac{dy}{dx} = 3x^2$
$\displaystyle 2x^2y \cdot \frac{dy}{dx} + 2 \cdot \frac{dy}{dx} = 3x^2 - 2xy^2$
$\displaystyle \frac{dy}{dx}(2x^2y + 2) = 3x^2 - 2xy^2$
$\displaystyle \frac{dy}{dx} = \frac{3x^2 - 2xy^2}{2x^2y + 2}$