Implicit differentiation to find dy/dx at a point

Differentiate both sides of $x^3 - y^3 = 6xy$ with respect to $x$. The left side gives $3x^2 - 3y^2 \frac{dy}{dx}$. The right side requires the product rule since it is a product of $x$ and $y$: $\frac{d}{dx}[6xy] = 6y + 6x\frac{dy}{dx}$. So the equation becomes $3x^2 - 3y^2 \frac{dy}{dx} = 6y + 6x\frac{dy}{dx}$.

Collect all terms with $\frac{dy}{dx}$ on one side: $-3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2$, so $-(3y^2+6x)\frac{dy}{dx} = 6y-3x^2$, giving $\frac{dy}{dx} = \frac{3x^2 - 6y}{3y^2+6x} = \frac{x^2-2y}{y^2+2x}$. Substituting $x=3, y=3$: $\frac{dy}{dx} = \frac{9-6}{9+6} = \frac{3}{15} = \frac{1}{5}$.

The right-hand side term six x y is a product of the two variable expressions x and y, so the product rule must be applied to differentiate it. This gives six y plus six x times dy over dx. Applying the quotient rule instead would be incorrect since there is no division here. The correct answer is the product rule.