Using Georgi's book on Lie algebras; one question asks to decompose the tensor product of the $3\otimes 3$ representation for $SU(3)$, which becomes $6 \oplus \bar{3}$. I understand how to do this using tensor representation, ie. writing $u^i v^j = k^{ij}$ where $k^{ij}$ is a sum of tensors that transform like $6$ and $\bar{3}$. But up to this point he has done diagrammatic construction with the highest weight method:
For example: Consider just the single $3 = (1,0)$, where $(1,0) = (q_1-p_1,q_2-p_2)$, and $q_i$ represents how many times you may lower with root $\alpha_i$ until $T_{-\alpha_i}v = 0$, and the same for $p_i$ with raising. This representation, through lowering $\alpha_1$ and $\alpha_2$, turns into $(1,0) \rightarrow (-1,1) \rightarrow (0,-1)$. I can't really draw this the way he does in the book.
I've managed to construct how the diagram should look for the tensor product $3 \otimes 3$, but I can't see how you would find the decomposition $6 \oplus \bar{3}$ from this.
EDIT: SOLVED
If you construct the diagram for $3 \otimes 3$ (for those looking for help on this same question, try making a diagram like in the book, but each entry in it with 2 boxes, one for lowering from the first $3$ and the other for the second), you obtain 9 entries chained together. One chain in the diagram looks like the diagram for the $6$ if you treat each $q-p$ pair as a new single $q-p$ value. The remaining three entries form a chain that looks like the $\bar{3}$.
ie. For the highest weight of $6$, $(1 \quad 0)(1 \quad 0) \rightarrow (2 \quad 0)$. For the highest weight of $\bar{3}$, $(1 \quad 0)(-1 \quad 1) \rightarrow (0 \quad 1)$.