Implicit differentiation and slope at a point

Differentiate both sides of $x^3 + y^3 - 3xy = 6$ with respect to $x$. The left side gives $3x^2 + 3y^2\frac{dy}{dx} - 3\left(y + x\frac{dy}{dx}\right)$, using the power rule on $x^3$ and $y^3$ (with the chain rule for $y^3$) and the product rule on $3xy$. The right side derivative is $0$ since $6$ is a constant.

So we have $3x^2 + 3y^2\frac{dy}{dx} - 3y - 3x\frac{dy}{dx} = 0$. Collecting the $\frac{dy}{dx}$ terms: $(3y^2 - 3x)\frac{dy}{dx} = 3y - 3x^2$, so $\frac{dy}{dx} = \dfrac{3y - 3x^2}{3y^2 - 3x} = \dfrac{y - x^2}{y^2 - x}$. Substituting $x=3$, $y=-1$ gives $\frac{dy}{dx} = \dfrac{-1 - 9}{1 - 3} = \dfrac{-10}{-2} = 5$.

The product rule was used on the term negative three x y, since it is a product of two functions of x, namely x and y. Differentiating it gave y plus x times dy dx, then multiplied by negative three. This is distinct from the chain rule used on y cubed.