= Solution
The Hessian determinant of $u^3+v^3+w^3$ is a nonzero scalar multiple of $uvw$. Hence the <inflection point of a plane cubic>[inflection points] are
$$
(1:-\zeta:0),
\qquad(1:0:-\zeta),
\qquad(0:1:-\zeta),
\qquad \zeta^3=1.
$$
There are nine of them. Since $O_C=(1:-1:0)$ is an inflection point, these are exactly $C[3]$. They are all defined over
$$
K=\mathbb Q(\zeta_3)=\mathbb Q(\sqrt{-3}),
$$
so $C(K)[3]\cong(\mathbb Z/3\mathbb Z)^2$.
Solved by gpt-5.6-sol high.
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