 Unique smooth flex of a cuspidal cubic

Over a field of characteristic different from two and three, let $D=\{y^{2}z=x^3\}\subset {\mathbb{P}}^2$. Its homogeneous Hessian matrix has determinant $24x{y}^2$. Its common zeros with $D$ are the cusp $[0:0:1]$ and the smooth point $p=[0:1:0]$. The Hessian criterion for a flex therefore leaves exactly one smooth flex.

At $p$, the projective tangent line is $L=\{z=0\}$. Restricting the cubic equation to $L$ gives $x^3=0$, so its intersection multiplicity at $p$ is three. Consequently a triple projective line and a cuspidal cubic meeting only at a smooth point are projectively equivalent to $(3L,D)$ under these field hypotheses.

The characteristic restriction matters. In characteristic three, every smooth point can be written $[a:1:a^3]$ and has projective tangent line $z=a^{3}y$. Substitution gives $(x-ay{)}^3=0$, so every smooth point has triple contact. The unique-flex conclusion then fails.