 Solution

On the nonsingular cubic, $x^3+y^3+z^3=-3\lambda xyz$. Substituting this into the Hessian equation leaves $-4({\lambda }^3+1)xyz=0$. Since ${\lambda }^3+1\ne 0$, a flex has a zero coordinate. The remaining cubic equation then gives the nine pointswhere $\omega =e^{2\pi i/3}$. All are distinct, smooth, and lie on both the cubic and its Hessian curve, so they are exactly the inflection points. In particular the nine flexes are independent of $\lambda$ within the nonsingular family. This is the flex configuration of the Hesse pencil.