Group of an Elliptic curve.
The computation is not so complicated.
We write $X = \frac x z, Y = \frac y z$ for the affine coordinates.
By doubling formula, we have $X(2P) = \frac{X^4 - 2X^2 + 8X + 1}{4(X^3 + X - 1)}$.
Now let $A$ be the point $(0, 2)$. Plug the value $X = 0$ into the doubling formula, we see that $X(2A) = 1$. This shows that the group $E(\Bbb F_5)$ cannot be isomorphic to $C_3 \times C_3$: otherwise we would have $2A = -A = (0, 3)$.
Therefore we have $E(\Bbb F_5) \simeq C_9$.