Example 7 explains that there are only three distinct points $A, B, C$ on the curve $x^{3}+3 x y+y^{3}=1$ that can form an equilateral triangle, and find the area of this equilateral triangle.
From formula (1), we know that $x^{3}+3 x y+y^{3}-1=0$ is actually reducible.
Because the left side of the equation can be factored as
$$
(x+y-1)\left(x^{2}-x y+y^{2}+x+y+1\right) \text {. }
$$

From formula (1), we know that the second factor can be written as
$$
\frac{1}{2}\left[(x+1)^{2}+(y+1)^{2}+(x-y)^{2}\right] \text {. }
$$

Thus, it is only zero at $(-1,-1)$. Therefore, the curve in question consists only of the single point $(-1,-1)$ and the line $x+y=1$.

To form a triangle with three points on this curve, one vertex must be $(-1,-1)$, and the other two vertices must lie on the line $x+y=1$. Thus, the altitude from $(-1,-1)$ is the distance from $(-1,-1)$ to $\left(\frac{1}{2}, \frac{1}{2}\right)$, which equals $\frac{3 \sqrt{2}}{2}$.

It is easy to know that the area of an equilateral triangle with height $h$ is $\frac{\sqrt{3}}{3} h^{2}$, so the required area is $\frac{3 \sqrt{3}}{2}$.