Answer
$$2x+6y^2\frac{dy}{dx}=0.$$
$$\frac{dy}{dx}=-\frac{x}{3y^2}.$$
$$\frac{dy}{dx}|_{(2,1)}=-\frac{2}{3}.$$
Work Step by Step
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
$$\frac{dy}{dx}|_{(2,1)}=-\frac{2}{3}.$$
