What is the mistake in finding the irreps of $SU(3)$ multiplets $6 \otimes 8$, or $15 \otimes \bar{15}$? I wrote a Mathematica paclet that can be used to find irreducible representations of $SU(n)$.
Specifically, given a product of $SU(n)$ multiplets, it can compute the corresponding sum.
To test the paclet, I compare its output with various examples from books. 
It fails on two occasions, when I compare against Table 4.13.1 on page 83 in The Lie Algebras of su(N) by Walter Pfeifer. 
I also did the calculations by hand, but I obtain the same answer as my program. 
My calculation:
$$6 \otimes 8 = \color{red}{\bar{3}} \oplus \bar{15} \oplus 6 \oplus 24$$
Book:
$$6 \otimes 8 = \color{red}{3} \oplus \bar{15} \oplus 6 \oplus 24$$
My calculation:
$$15 \otimes \bar{15} = 1 \oplus \bar{10} \oplus \color{red}{10} \oplus 8 \oplus 8 \oplus \bar{35} \oplus 27 \oplus 27 \oplus 64 \oplus 35$$
Book:
$$15 \otimes \bar{15} = 1 \oplus \bar{10} \oplus \color{red}{\bar{10}} \oplus 8 \oplus 8 \oplus \bar{35} \oplus 27 \oplus 27 \oplus 64 \oplus 35$$
Is this my mistake, or are these typos in the book?
Here is the explicit calculation of the first example, using Young tableaux:

 A: Following the recipe for the multiplicities of components of a tensor product outline by Jim Humphreys here. IIRC this recipe is a consequence of Weyl's character formula. See Humphreys' book for a proof. In my edition this is Exercise 8 in section 24.4 (Steinberg's formula)
Assuming that the fundamental dominant weights are indexed in such a way that the representation you denote by $6$ is $V(2\lambda_1)$. In the case of the adjoint representation there is no ambiguity and the 8-dimensional representation is the one denoted by $V(\lambda_1+\lambda_2)$.
The module $V(2\lambda_1)$ has six distinct weights, each with multiplicity one, namele $2\lambda_1, \lambda_2, -2\lambda_1+2\lambda_2, \lambda_1-\lambda_2,-\lambda_1$ and $-2\lambda_2$. Therefore in the tensor product we get summands with highest weights (add $\lambda_1+\lambda_2$ to the weights listed above): $3\lambda_1+\lambda_2$,$\lambda_1+2\lambda_2$, $-\lambda_1+3\lambda_2$, $2\lambda_1$, $\lambda_2$ and $\lambda_1-\lambda_2$. Among these six weight the non-dominant ones have a non-trivial stabilizer under the dot action, so we simply throw those out. The result is that
$$
V(2\lambda_1)\otimes V(\lambda_1+\lambda_2)\simeq V(3\lambda_1+\lambda_2)\oplus V(\lambda_1+2\lambda_2)\oplus V(2\lambda_1)\oplus V(\lambda_2).
$$
So the same six-dimensional rep $V(2\lambda_1)$ we used as a factor in the tensor product also appears as a summand. The module $V(\lambda_2)$ is the dual of the 3-dimensional module $V(\lambda_1)$, and I'm fairly sure that physicists denote it $\overline{3}$. This is good news in the sense that it says that you're right and your other source is wrong!
Of the other two summands Weyl's dimension formula says that
$$
\dim V(3\lambda_1+\lambda_2)=\frac{4\cdot6\cdot2}{1\cdot2\cdot1}=24,
$$
and
$$
\dim V(\lambda_1+2\lambda_2)=\frac{2\cdot5\cdot3}{1\cdot2\cdot1}=15.
$$
The latter module is the dual of $V(2\lambda_1+\lambda_2)$, obviously also 15-dimensional. I may be wrong, but IIRC the way physicists' notation plays out is that of the non-self-dual modules of $SU(3)$ the one whose highest weight $m_1\lambda_1+m_2\lambda_2$ satisfies the inequality $m_1>m_2$ is denoted without the overline, while the other (with highest weight $m_2\lambda_1+m_1\lambda_2$) gets the "bar".
Assuming this recollection is in line with the physicist's notation, we get, in the end
$$
6\otimes 8\simeq 24\oplus6\oplus\overline{15}\oplus\overline{3}.
$$

Another argument leading to the same conclusion is that the weights of $3=V(\lambda_1)$ are in a different coset of the root lattice from the rest of the weights appearing in this decomposition. Therefore $\overline{3}=V(\lambda_2)$ can appear (and actually does appear) as a composition factor, but $3$ cannot.
