1. **State the problem:** Given the curve defined by the equation $$x^3 + y^3 = 6xy$$, find the derivative $$\frac{dy}{dx}$$ using implicit differentiation. 2. **Recall the formula and rules:** To differentiate implicitly, differentiate both sides of the equation with respect to $$x$$, treating $$y$$ as a function of $$x$$ (so use the chain rule for terms involving $$y$$). 3. **Differentiate each term:** - $$\frac{d}{dx}(x^3) = 3x^2$$ - $$\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$$ (chain rule) - $$\frac{d}{dx}(6xy) = 6 \left( x \frac{dy}{dx} + y \right)$$ (product rule) 4. **Write the differentiated equation:** $$3x^2 + 3y^2 \frac{dy}{dx} = 6 \left( x \frac{dy}{dx} + y \right)$$ 5. **Group terms with $$\frac{dy}{dx}$$ on one side:** $$3y^2 \frac{dy}{dx} - 6x \frac{dy}{dx} = 6y - 3x^2$$ 6. **Factor out $$\frac{dy}{dx}$$:** $$\frac{dy}{dx} (3y^2 - 6x) = 6y - 3x^2$$ 7. **Solve for $$\frac{dy}{dx}$$:** $$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x}$$ 8. **Simplify the fraction by dividing numerator and denominator by 3:** $$\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}$$ **Final answer:** $$\boxed{\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}}$$