Clebsch-Gordan Coefficients for 8 in $3 \otimes \bar 3 $ and the $6$ in $3 \otimes 3$ of $SU(3)$? Do Tables for the Clebsch Gordan coefficients for the decomposition of the $8$ dimensional irrep of $SU(3)$ into $3 \otimes \bar 3 $ and the $6$ in $3 \otimes 3$ (in the Dynkin basis) exist somewhere? (Of course, I googled and searched on my own for quite a while, but was unable to find anything useful)
I tried to compute them by hand, but I'm not entirely sure when it comes to normalization of the states. Especially, lowering the degenerate $(0,0)$ state is problematic, because there are four possibilities, which lead to two different states, but with different numerical coefficients. 
I tried to compute them using a software called CleGo, which gave me 
[[[("1", ("(0,-1,)1", "(-1,0,)1"))]];
[[("2", ("(0,-1,)1", "(1,-1,)1"))]; [("1", ("(-1,1,)1", "(-1,0,)1"))]];
[[("1", ("(-1,1,)1", "(1,-1,)1")); ("1", ("(1,0,)1", "(-1,0,)1"))];
  [("1", ("(0,-1,)1", "(0,1,)1")); ("1", ("(-1,1,)1", "(1,-1,)1"))]];
[[("1", ("(1,0,)1", "(1,-1,)1"))]; [("1", ("(-1,1,)1", "(0,1,)1"))]];
[[("1", ("(1,0,)1", "(0,1,)1"))]]]
and I'm not sure how to interpret this. 
 A: The calculations are reasonably straightforward for your specific cases.
The simplest approach is to consider the states in $(1,0)$ to be the 1-excitations states of the 3D harmonic oscillator: $\{\vert 100\rangle,\vert 010\rangle,\vert 001\rangle\}$.   For a state $\vert n_1n_2n_3\rangle$, the weight is $[n_1-n_2,n_2-n_3]$.  There is no weight multiplicity in $(1,0)$ so $(n_1,n_2,n_3)$ uniquely label a basis in the irrep. 
The states in $(2,0)$ must then, by Schur-Weyl duality, be in the symmetric part of $(1,0)\otimes(1,0)$.  It follows immediately by additivity of weights and permutation symmetry that 
\begin{align}
\vert 200\rangle &= \vert 100\rangle\vert 100\rangle \, ,\\
\vert 110\rangle &= \frac{1}{\sqrt{2}}
\left(\vert 100\rangle\vert 010\rangle+\vert 010\rangle\vert 100\rangle\right)
\end{align}
and so forth for the other states in $(2,0)$.
The case of $(1,0)\otimes (0,1)$ can be handled in the same general manner, provided you construct the states in $(0,1)$ as the antisymmetric part of $(1,0)\otimes (0,1)$.   You can then construct states in $(1,1)$ as 3-particle states.
Thus, for instance, the highest weight state of $(0,1)$ is
$$
\vert(0,1)110\rangle = \frac{1}{\sqrt{2}}
\left(\vert 100\rangle \vert 010\rangle - \vert 010\rangle \vert 110\rangle\right)
$$
and the highest weight state of $(1,1)$ will then be 
$$
\vert (1,1)210\rangle = \vert 100\rangle \vert (0,1)110\rangle. \tag{1}
$$
The irrep (1,1) has 2 states of weight $0$.  It is not hard to find they are of the form
$$
\vert\psi\rangle = a\vert 100\rangle \vert(01)011\rangle + b
\vert 010\rangle \vert(01)\vert 101\rangle + c\vert 001\rangle \vert (01)\vert 110\rangle \tag{2}
$$
These must be orthogonal to the singlet state in $(0,0)$.  This singlet state is in fact the determinant of the three-particle states
$$
\vert (0,0)111\rangle =
\left(\begin{array}{ccc}
\vert 100\rangle & \vert 010\rangle &\vert 001\rangle \\
\vert 010\rangle &\vert 001\rangle &\vert 100\rangle \\
\vert 001\rangle & \vert 100\rangle &\vert 010\rangle
\end{array}\right) \tag{3}
$$
The zero-weight states in (1,1) must be orthogonal to (3).  If you use the standard $su(2)\oplus u(1)$ basis, you can construct a state with $J=0$ by forcing the coefficients in (2) to be give you a null state when acting on (2) using any operator in the $su(2)$ subalgebra in addition to being orthogonal to (3).  The last state with zero-weight is a $J=1$ state.  You can construct it by orthgonality to the $J=0$ state of $(1,1)$ and the $(0,0)$ state.
All other states are uniquely labelled by weights so they have a form,  in terms of $(1,0)\otimes(0,1)$ states, obtained by suitable modification of (1).
This wikipage gives additional references.  There is also this online CG calculator which might be useful.  As always, the difficulties are with the notation and the relative phases of the multiplets (unlike the $su(2)$ case, there is nothing like the Condon-Shortley phase convention in $su(3)$, and this has caused considerable pain to external users of a specific convention.)
