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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.

Choose homogeneous polynomials $F,G$ defining the two projective plane curves. Their absence of a common irreducible component makes them coprime. For a point $O=[1:0:0]$ outside both curves, they have degrees $n,m$ in $X$ and nonzero constant leading coefficients. The resultantis therefore nonzero. Indeed, a vanishing resultant over $k(Y,Z)$ would give a common factor there, and Gauss's lemma would give a common component of the original curves.

The homogeneity of $F,G$ makes $R$ a homogeneous polynomial of degree $nm$: scaling $(Y,Z)$ by $a$ scales their $X$-roots by $a$, and the product of the $nm$ differences of roots scales by $a^{nm}$. Every point in the intersection projects from $O$ to a zero of $R$ in the projective line. Each such projection line contains only finitely many points of either curve, since it passes through $O$ outside them. Thus the intersection is finite already.

We may now choose $O$ also outside the finitely many joining lines of distinct intersection points. Projection is then injective on the intersection. A nonzero degree-$nm$ homogeneous polynomial on the projective line has at most $nm$ distinct zeros, so the resultant bound for intersections of plane curves givesThis derives the required cardinality bound without assuming the full intersection-multiplicity statement of Bézout's theorem.

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.