Question Detail

We use implicit differentiation with respect to (x):(\frac{d}{dx}(x^3) + \frac{d}{dx}(x) + \frac{d}{dx}(y^3) + \frac{d}{dx}(3y) = \frac{d}{dx}(6))(3x^2 + 1 + 3y^2\frac{dy}{dx} + 3\frac{dy}{dx} = 0)Factor out (\frac{dy}{dx}):((3y^2 + 3)\frac{dy}{dx} = -(3x^2 + 1))(\frac{dy}{dx} = -\frac{3x^2 + 1}{3y^2 + 3})Now, substitute the point ((x, y) = (1, 1)):(\frac{dy}{dx} = -\frac{3(1)^2 + 1}{3(1)^2 + 3})(= -\frac{3 + 1}{3 + 3})(= -\frac{4}{6})(= -\frac{2}{3}).