I'm coming from a projective setting of a smooth cubic plane curve over a field $K$ and want to show that I can bring it to the Weierstrass long form. The usual method is to start with a point and bring it to $(1:0:0)$ and then take the tangent line of this point and bring it to $z=0$ without changing $(1:0:0)$. But when doing this, one wants that the tangent line of the chosen point meet the curve at another point. I want to argue why this can happen at all.
In other words, I want to argue why a smooth cubic curve in the projective plane will have a point whose tangent line does not meet the point with multiplicity $3$. So if $C$ is the curve and $P$ any point and I denote $L_P$ as the tangent line at that point, do I really need to use commutative algebra and show that there is a $P$ such that $I(C\cap L_P , P) =2$ or is there a more elementary method of showing this? Here, when I meant commutative algebra, I view the intersection number as the $K$ vector space dimension of the polynomial ring localized at $P$ modulo the defining equations for the curve and the tangent line.
Edit: From the answer and comments, I realize that I should probably use case distinctions if I want to show that I can bring it into Weierstrass long form (by some isomorphisms). Namely, cases when I have only a flex (how to show in this case the curve consist of only finite number of points?) and cases when there is a non-flex. I suppose I can argue that there is a non-flex over $\bar K$ (algebraic closure), but I think I should resist this because I still want to have this curve defined over $K$ and fix the $K$-rational points.