Every smooth cubic curve has a flex point

Question

I want to show that every smooth irreducible plane cubic $C$ has a flex point, i.e. a point $P$ with $i_P(C, T_C(P)) = 3)$ (where $T_C(P)$ is the tangent to $C$ at $P$). I know how to do this in characteristic zero - take the Hessian and use Bezout's theorem - but don't know how to proceed in characteristic $p$. Can anyone provide a proof or point me to a resource where this is shown? I'm fine with chracteristic $\neq 2,3$.