## Calculus (3rd Edition)

$y' = -\frac{2x}{9y^2}$
$\frac{d}{dx}(3y^3) + \frac{d}{dx}(x^2) = \frac{d}{dx}(5)$ $3*3y^2(y') + 2x = 0$ $9y^2(y') + 2x = 0$ Now we have to solve for $y'$: $9y^2(y') = -2x$ $y' = -\frac{2x}{9y^2}$