Let $C=\{(3,1),(2,1)\}$ be basis for $\mathbb{R}^2$.How to find coordinates for trilinear form $H=e^1\bigotimes e^1\bigotimes e^2 - e^2\bigotimes e^2\bigotimes e^1$ relative to $C$ and canonical basis.
$e^1,e^2,e^3$ are elements of a dual canonical basis.
Your new basis is written $$b_1=3e_1+e_2,$$ $$b_2=2e_1+e_2.$$
Solving for $e_1,e_2$ $$e_1=b_1-b_2,$$ $$e_2=-b_1+3b_2.$$
Since one wants $e^i(e_j)=\delta^i_j$, the duals for $b_i$ are $$b^1=e^1-2e^2,$$ $$b^1=-e^1+3e^2.$$ Then one can check that $b^i(b_j)=\delta^i_j$. Solving for $e^1,e^2$ will give $$e^1=3b^1+2b^2,$$ $$e^2=b^1+b^2.$$
Subbing on the terms of $H$ gives $$e^1\otimes e^1\otimes e^2=(3b^1+2b^2)\otimes(3b^1+2b^2)\otimes(b^1+b^2),$$ and $$e^2\otimes e^2\otimes e^1=(b^1+b^2)\otimes(b^1+b^2)\otimes(3b^1+2b^2),$$ that can be simplified by the use of distributive properties, then to assemble your $H$.
Can you take it from here?
Addendum:
Let me illustrate more with $e^1\otimes e^1\otimes e^2$. This is $$=e^1\otimes(3b^1\otimes b^1+3b^1\otimes b^2+2b^2\otimes b^1+2b^2\otimes b^2),$$
$$=(3b^1+2b^2)\otimes(3b^1\otimes b^1+3b^1\otimes b^2+2b^2\otimes b^1+2b^2\otimes b^2),$$
$$=9b^1\otimes b^1\otimes b^1+ 9b^1\otimes b^1\otimes b^2+ 6b^1\otimes b^2\otimes b^1+ 6b^1\otimes b^2\otimes b^2+ 6b^2\otimes b^1\otimes b^1+\\ 6b^2\otimes b^1\otimes b^2+ 4b^2\otimes b^2\otimes b^1+ 4b^2\otimes b^2\otimes b^2.$$
I hope this helps.