I reason as follows:

If $P$ is a point in an elliptic curve $E$ with order 3, then $2P = -P$. For any point $P$ that satisfies this relation, we have $2(-P) = P$, ie the reflection also satifies.

Let a cubic curve be given by $y^2 = x^3 + b_3x^2+b_2x + b_0$, and let $P = (x_1,y_1)$ and $2P = (x_3,-y_3)$. By our relation we must have $(x_3,-y_3)= (x_1,-y_1)$ so $x_3 = x_1$. The addition formula for $2P$ explicitly gives $x_3$ by $$x_3 = (\frac{3x_1^2+2b_2x_1+b_1}{2y_1})^2 - b_2 - 2x_1$$ and setting $x_3 = x_1$ gives

$$0 = (\frac{3x_1^2+2b_2x_1+b_1}{2y_1})^2 - b_2 - 3x_1$$

This is a fourth degree polynomial in $x_1$, and thus has at most 4 real solutions, and so at most 4 rational solutions. We thus get at most 4 possible candidates for $x_1$ for a given cubic curve, and by symmetry we get 4 more. Thus there are at most 8 possible coordinates for $P$.

Note: I did not elaborate on $y_3$, since the addition formula that specifies $y_3$ is simply equal to $y_1$ if $x_3 = x_1$, so for any $x_o$ that satisfies the above relations $P = (x_o,y)$ satisfies for all $y$.

I'm just learning about elliptic curves and would appreciate any and all feedback on this solution!

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    $\begingroup$ A good start. But you should eliminate $y_1$ from that equation. Otherwise you cannot conclude that there are only four possibilities for $x$, because with $y$ still in the equation, you get at most four possibilities for $x$ for each $y$. $\endgroup$ – Jyrki Lahtonen Oct 19 '16 at 9:26
  • $\begingroup$ I would use an entirely different approach, since the points of order three are precisely the inflection points. As I recall, when you write down the condition $\frac{d^2y}{dx^2}=0$, the eightfoldness of the solution set is apparent. $\endgroup$ – Lubin Oct 26 '16 at 14:23
  • $\begingroup$ On whether the points of order three all might be real, No Way! It’s fancier stuff than you’ve seen but the $e_3$-pairing on points of order three is into the cube roots of unity. So if all are $k$-rational, then $k\supset\Bbb Q(\omega)$, where $\omega=\frac12(-1+\sqrt{-3}\,)$. $\endgroup$ – Lubin Oct 26 '16 at 14:30

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