Let $E\subset\Bbb{P}_\Bbb{C}^2$ be a smooth cubic with a flex $O\in E$. In particular, there is a group structure $(E,\oplus)$ with $O$ as identity element induced by a bijection $E\to\text{Pic}(E)$ given by $P\mapsto P-O$.
It is a classical fact that $P_1\oplus P_2\oplus P_3=O$ implies that $P_1,P_2,P_3$ are in a same line.
My question is: is it true that $P_1\oplus P_2\oplus...\oplus P_6=O$ implies that $P_1,P_2,...,P_6$ are in a same conic?
Here is my attempt to prove it: if $P_1\oplus P_2\oplus...\oplus P_6=O$, then in $\text{Pic}(E)$ we get $$(P_1+P_2+...+P_6)-6\cdot O=0$$
Therefore we have some $f\in k(E)$ having $P_1,...,P_6$ as zeros and $O$ as a pole with order 6. Let $\ell\in k[x_0,x_1,x_2]$ be the linear form defining the line tangent to $E$ at $O$. Then $f=(q/\ell^2)|_E$, where $q\in k[x_0,x_1,x_2]$ is homogeneous of degree 2.
The last sentence looks very convincing, but I'm not sure if I can do that. Can we say that $f$ is the restriction of some quotient of homogeneous polynomials?
Thank you!