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For a cubic homogeneous polynomial $F$, its Hessian matrix has linear entries, so its determinant is a cubic homogeneous polynomial, possibly zero. On a nonsingular plane cubic in characteristic different from two, its zeros are exactly the flexes, by the Hessian criterion for a flex. If the determinant is nonzero, Bézout's theorem supplies intersections; if it is zero, the local criterion itself supplies flexes.

Let $F$ define the nonsingular plane cubic and let $H_F=\det \mathrm{Hess}F$ be its Hessian curve of a plane cubic equation. The Hessian criterion for a flex applies in the allowed field characteristic. Here is the local calculation, so no characteristic-zero argument is needed.

Move any point $P$ to $[0:1:0]$ and its tangent line to $Z=0$. Smoothness gives a nonzero coefficient $g$ of $Y^{2}Z$, while the coefficients of $Y^3$ and $X{Y}^2$ vanish. ThusThe intersection multiplicity with the tangent line is at least three precisely when $b=0$, which is precisely $H_F(P)=0$ because the field characteristic is not two. The tangent cannot be a component of a nonsingular plane cubic.

If $H_F$ vanishes identically, every point is a flex by this calculation. Otherwise it is a degree-three homogeneous polynomial. Its zero locus meets the cubic by Bézout's theorem; if they share a component, that also supplies a point of intersection. Every such point is smooth and satisfies the criterion, so a flex exists.

Apply a projective transformation to put the cuspidal cubic in the form $D=\{F=y^{2}z-x^3=0\}$. Since $L$ meets $D$ at only one smooth point $p$, the Bézout theorem forces intersection multiplicity three there. Hence $L$ is the tangent line at a flex.

The homogeneous Hessian matrix has determinantBy the Hessian criterion for a flex, every smooth flex lies in $D\cap \{x{y}^2=0\}$. These intersections are $[0:0:1]$, the cusp, and $[0:1:0]$, which is smooth. Thus the unique smooth flex of a cuspidal cubic is $p=[0:1:0]$, with tangent line $L=\{z=0\}$. Conversely the pair in part (b) has exactly the required properties. Therefore there is one projective-equivalence class, represented by

For $D=\{y^{2}z=x^3\}$, the determinant of the homogeneous Hessian matrix is $24x{y}^2$. Its intersections with $D$ are the cusp $[0:0:1]$ and the smooth point $[0:1:0]$. The Hessian criterion for a flex therefore gives a unique smooth flex, with tangent $z=0$ and contact order three. Consequently every pair consisting of a triple line and a cuspidal cubic meeting only at a smooth point is projectively equivalent to $(3\{z=0\},D)$.