Loop the Loop

Let us differentiate this equation with respect to $x$ to give:

$3x^2 + 3y^2 y' = 6y + 6xy' \Rightarrow y' = \frac{6y - 3x^2}{3y^2 - 6x}$ (i).

If a horizontal tangent exists at the first-quadrant point $(A,B)$ , then solving for the point which makes (i) equal to zero yields:

$y' = \frac{6y - 3x^2}{3y^2 - 6x} = 0 \Rightarrow y = \frac{1}{2} x^2$ (ii).

or $(A,B) = (k, \frac{1}{2} k^2)$ for $k \in \mathbb{R}^{+}.$ Substituting this point into our original equation will allow us to solve for $k^3$ since $A \times B = \frac{1}{2} k^3$ :

$k^3 + (\frac{1}{2} k^2)^{3} = 6k \cdot \frac{1}{2} k^2 \Rightarrow k^6 - 16k^3 = 0 \Rightarrow k^3 ( k^3 - 16) = 0 \Rightarrow k^3 = 0, 16$

with $k^3 = 16$ being the only satisfactory value for the first quadrant. Hence, $A \times B = \frac{1}{2} \cdot 16 = \boxed{8}.$

This is a specific form of Folium of Descartes

Solve it like this :

The coordinate is $(2 \sqrt[3]{2}, 2 \sqrt[3]{4} )$ , and their product is simply $4 \cdot \sqrt[3]{8} = \boxed{8}$