= Solution
The <Riemann-Roch theorem> states
$$
\ell(D)-\ell(K-D)=\deg D+1-g.
$$
On a genus-one curve $K\sim0$. If $D$ has degree zero, then $D+P_0$ has degree one, and Riemann--Roch gives $\ell(D+P_0)=1$. Its unique nonzero section has an effective divisor of degree one, say $P$, so
$$
D\sim P-P_0.
$$
If also $D\sim Q-P_0$, then $P\sim Q$; a nonconstant <function> with divisor $P-Q$ would define a degree-one map to $\mathbb P^1$, forcing the genus to be zero. Hence $P=Q$.
This identifies $X$ with $\operatorname{Pic}^0(X)$ by $P\mapsto[P-P_0]$ and defines $P+Q=R$ through
$$
[P-P_0]+[Q-P_0]=[R-P_0].
$$
For the stated cubic, the line through $A=(2,1)$ and $B=(-2,-1)$ is $y=x/2$. Its third intersection is $(1/4,1/8)$, so reflection in the $x$-axis gives
$$
A+B=(1/4:-1/8:1).
$$
Because $P_0$ is an inflection point, a line section is linearly equivalent to $3P_0$. A point $P$ is an inflection point exactly when some line has intersection divisor $3P$, equivalently
$$
3P\sim3P_0.
$$
Under the group-law identification this is precisely $3(P-P_0)=0$, so the inflection points are exactly the three-torsion points.
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