## Calculus (3rd Edition)

$$2x+6y^2\frac{dy}{dx}=0.$$ $$\frac{dy}{dx}=-\frac{x}{3y^2}.$$ $$\frac{dy}{dx}|_{(2,1)}=-\frac{2}{3}.$$
Differentiating the equation $x^2+2y^3=6$ with respect to $x$ and using the chain rule, we get $$2x+6y^2\frac{dy}{dx}=0.$$ Hence, $$\frac{dy}{dx}=-\frac{x}{3y^2}.$$ Moreover, we have $$\frac{dy}{dx}|_{(2,1)}=-\frac{2}{3}.$$