## Calculus (3rd Edition)

$$2x+2xy'+2y+3y^2y'=0.$$
By using the product rule: $(uv)'=uv'+u'v$, we have $$2x+2xy'+2y+3y^2y'=0.$$ So the right answer is $$2x+2xy'+2y+3y^2y'=0.$$