1. **State the problem:** Differentiate implicitly the equation $$x^3 + y^3 = 6xy$$ with respect to $x$ to find $\frac{dy}{dx}$. 2. **Recall the rules:** - Use the power rule: $$\frac{d}{dx}[x^n] = nx^{n-1}$$. - Use implicit differentiation: treat $y$ as a function of $x$, so $$\frac{d}{dx}[y^3] = 3y^2 \frac{dy}{dx}$$. - Use the product rule for $$6xy$$: $$\frac{d}{dx}[6xy] = 6\left(y + x\frac{dy}{dx}\right)$$. 3. **Differentiate both sides:** $$\frac{d}{dx}[x^3] + \frac{d}{dx}[y^3] = \frac{d}{dx}[6xy]$$ $$3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx}$$ 4. **Group terms with $$\frac{dy}{dx}$$ on one side:** $$3y^2 \frac{dy}{dx} - 6x \frac{dy}{dx} = 6y - 3x^2$$ 5. **Factor out $$\frac{dy}{dx}$$:** $$\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2$$ 6. **Solve for $$\frac{dy}{dx}$$:** $$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x}$$ 7. **Simplify the fraction by dividing numerator and denominator by 3:** $$\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}$$ **Final answer:** $$\boxed{\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}}$$