Implicit differentiation of the folium of Descartes

Differentiate both sides of $x^3 + y^3 = 9xy$ with respect to $x$. On the left side, $\frac{d}{dx}[x^3] = 3x^2$ and $\frac{d}{dx}[y^3] = 3y^2 \frac{dy}{dx}$ by the chain rule since $y$ is a function of $x$. On the right side, apply the product rule to $9xy$: $\frac{d}{dx}[9xy] = 9\left(1\cdot y + x\cdot \frac{dy}{dx}\right) = 9y + 9x\frac{dy}{dx}$. This gives $3x^2 + 3y^2\frac{dy}{dx} = 9y + 9x\frac{dy}{dx}.$

Now collect all terms involving $\frac{dy}{dx}$ on one side: $3y^2\frac{dy}{dx} - 9x\frac{dy}{dx} = 9y - 3x^2$. Factor out $\frac{dy}{dx}$ on the left: $\frac{dy}{dx}(3y^2 - 9x) = 9y - 3x^2$. Dividing both sides by $(3y^2 - 9x)$ and simplifying by factoring out a 3 from numerator and denominator gives the final result.