In F. Calegari's lecture notes, Exercise.3.0.4 is: Show the $v_{3}(a_{2})<\frac{3}{4}$ being the sufficient condition for an elliptic curve having canonical order $3$ subgroup, where $v_{3}$ denotes a valuation of $\mathcal{O}_{K}$ s.t. $v_{3}(3)=1.$ Here the elliptic curve is defined over the ring of integers $\mathcal{O}_{K}$ with finite extension $K/\Bbb{Q}_{3}$ given by the minimal Weierstrass equation: $$y^{2}+a_{1}xy+a_{3}y=x^{3}+a_{2}x^{2}+a_{4}x+a_{6}.$$
I failed in solving this exercise.
The present question consists of the above part (any hints will be very welcomed) and the following:
Nevertheless, I found something weird while trying to solve this problem and I would like to ask why the following has happened (or say, what's wrong?): Assume now we dealing with an elliptic curve given by $F$ over $K/\Bbb{Q}_{3},$ with $$F(x,y,x_{2})=y^{2}x_{2}-x^{3}-Axx_{2}^{2}-Bx_{2}^{3}=0.$$ This enables us to apply Excercise.3.9 (Silverman. The arith. of ell. curve.) which claims that $P+P+P=O$ (the group structure of elliptic curve) iff the determinant of the Hessian (with $x_{0}:=x,$ $x_{1}:=y$) $$\text{det}\bigg(\frac{\partial^{2}F}{\partial x_{i}x_{j}}\bigg)_{0\leq i,j\leq 2}$$ equals $0.$ Require the determinant of Hessian vanishing at $P=(x,y,x_{2})$ and $P$ belonging to the curve given by $F.$ After some routine calculations, the following equation is deduced after $x_{2}=1$ is assumed: $$x^{4}+\frac{3}{2}Ax^{2}+\frac{5}{2}Bx-A^{3}/6=0.$$ Since $``2(v_{3}(A)+1)<v_{3}(A^{2})-1$ never happening $"$ indicates $4$ solutions of this equation never has different valuations, we seem to obtain the claim that order $3$ canonical subgroup does not exist in the elliptic curve given by $F,$ which looks wired. (See p.27 of F. Calegari's lecture for a similar argument).