From washington's book exercise 3.2. I have to show that $E[3] = \mathbb{Z}_3 \times \mathbb{Z}_3$ where $E$ is defined over a characteristic 2 field.
Given the generalized Weierstrauss equation $y^2 + xy = x^3 + a_2 x^2 + a_6$ (with $a_4 = 0$), then the point doubling formula is: $$x_2 = (x_0^4 + a_6) / x_0^2$$ Using the fact that $x(2P) = x(-P)$ for points in $E[3]$, we get that $$x^4 + x^3 + a_6 = 0$$ So there are 4 possible solution for $x$ in $K^\times$.
Now I just need to prove there are 2 possible values for $y$, and then including the infinity point, we will get a total of 9 points proving the group is isomorphic to $\mathbb{Z}_3 \times \mathbb{Z}_3$. This is where I become stuck.
The point doubling formula gives $y_2 = x_1 + y_1$ which after substituting gets: $$y_2 = x_1 + (y_0 + x_0^2)(x_1 - x_0)/x_0 + y_0$$ Please advise where to proceed from here. Thanks