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Put $F=x^3+y^3+13z^3-2xyz$. The Hessian criterion for flexes of a plane cubic applies because the projective cubic is smooth. To check smoothness, a singular point with all coordinates nonzero would satisfy $3x^2=2yz$, $3y^2=2xz$, $39z^2=2xy$. Multiplication would force $351=8$, a contradiction. If one coordinate is zero, the three equations force all coordinates to vanish, which is not a projective point.

For completeness, the Hessian criterion for flexes of a plane cubic follows locally by putting a smooth point at $(0:0:1)$ and its tangent at $y=0$. After normalization the homogeneous cubic has terms $y{z}^2+a{x}^{2}z+bxyz+c{y}^{2}z$ and terms cubic in $x,y$. Its tangent has contact order three precisely when $a=0$. Its Hessian matrix at the point has determinant $-8a$, giving exactly the same condition.

For the present cubic the Hessian determinant isOn $F=0$ this is $2744xyz=8\cdot 343xyz$. Thus the plane cubic flexes are exactly the points on the cubic with $xyz=0$. Let $c=\sqrt[3]{13}$ and let $\omega$ be a primitive cube root of unity. The complete list over the algebraic closure isThese nine points are distinct, and the Hessian calculation proves there are no others. They are the flexes of a Hesse cubic. The only rational one is $O^{\prime }=(1:-1:0)$: the other two points with $z=0$ require a nonrational cube root of unity, and a rational point in either remaining triple would give a rational cube root of $-13$. By prime valuations, $13$ is not a rational cube.