Does every irreducible projective cubic curve necessarily have a nonsingular point of inflection?
I've been trying to construct counterexamples, to no avail, which leads me to believe the question can be answered in the affirmative. Here we take $\mathbb{C}$ to be the ground field.
I think so. You can always stick an irreducible cubic into a more or less standard form $y^2=x(x-\lambda_1)(x-\lambda_2)$ by a linear change of variables. Then on the patch $y=1$ the curve is nonsingular -- either the whole curve is nonsingular or it has a unique singular point at the origin in the $xy$-plane, depending on the values of the $\lambda$s -- and in the form $z=x^3+ax^2z+bxz^2$. The tangent line here is clearly $z=0$ and it is also clearly inflectionary, as there is no degree-$2$ piece.