Maximum area of triangle on a curve

To tackle this one, we first "complete the cube" on the left side:

$\begin{aligned} x^3 + 3x^2y + 3xy^2 + y^3 &= 3x^2y + 3xy^2 - 3xy +1 \\ (x+y)^3 -1 &= 3x^2y + 3xy^2 - 3xy \\ (x + y - 1) \left[ (x + y)^2 + (x + y) + 1 \right] &= 3xy \ (x + y - 1) \\ (x + y - 1) (x^2 - xy + y^2 + x + y + 1) &= 0 \qquad \qquad \small \color{#3D99F6} \text{[Now multiply both sides by }2] \\ (x + y - 1) (2x^2 - 2xy + 2y^2 + 2x + 2y + 2) &= 0 \\ (x + y - 1) (x^2 - 2xy + y^2 + x^2 + 2x + 1 + y^2 + 2y + 1) &= 0 \\ (x + y - 1) \left[ (x - y)^2 + (x + 1)^2 + (y + 1)^2 \right] &= 0 \end{aligned}$

so either $(x + y - 1) = 0$ or $x = y = \text{-} 1$ . Thus the graph of $x^3 + 3xy + y^3 = 1$ consists of the line $x + y = 1$ plus the point $(\text{-}1,\text{-}1)$ , so there's a single equilateral triangle whose vertices all touch the graph.

Finally, from the graph, the distance from $(\text{-}1,\text{-}1)$ to the line $x + y + 1$ is easily seen to be $\dfrac{3\sqrt{2}}{2}$ , and this would be the height $h$ of the equilateral triangle in question. Then its area $A$ is

$A = \dfrac{\sqrt{3}}{3}h^2 = \dfrac{\sqrt{3}}{3} \left( \dfrac{3\sqrt{2}}{2} \right)^2 = \dfrac{\sqrt{3}}{3} \cdot \dfrac{9}{2} = \boxed{ \dfrac{3\sqrt{3}}{2} }$

By factorisation

$x^3+3xy+y^3-1=\frac{1}{2}(x+y-1)((x-y)^2+(y+1)^2+(x+1)^2))$

Therefore the set of points on the curve $x^3+3xy+y^3=1$ is the union of $(-1, -1)$ and the curve $y=-x+1$

This means that there is a unique equilateral triangle whose vertices are on the curve, and $(-1, -1)$ must be one of its vertices.

We can now easily compute the area of this triangle, which turns out to be $\displaystyle\frac{3\sqrt{3}}{2}$