= Solution
Let $V=\mathbb C^6$ be the defining <fundamental representation> of $SU(6)$. Its <weights> are $w_i=e_i-\tfrac16\sum_{j=1}^6e_j$. Pairing with the <simple coroots> gives <Dynkin labels> $(w_i)_a=\delta_{ia}-\delta_{i,a+1}$. Thus the complete list is
$$
\begin{array}{c|rrrrr}
&a_1&a_2&a_3&a_4&a_5\\\hline
w_1&1&0&0&0&0\\
w_2&-1&1&0&0&0\\
w_3&0&-1&1&0&0\\
w_4&0&0&-1&1&0\\
w_5&0&0&0&-1&1\\
w_6&0&0&0&0&-1
\end{array}.
$$
Successive <lowering operators> subtract the corresponding rows of the <Cartan matrix> in <4/a>. Each <weight multiplicity> is one.
The complex-conjugate <representation> $\overline{\mathbf6}\cong V^*$ is the <dual representation> and has the negatives of these <weights>. Its <highest weight> is $-w_6=(0,0,0,0,1)$, and its entire <weight decomposition> is
$$
\boxed{\operatorname{Wt}(\overline{\mathbf6})=\left\{
\begin{array}{r}
(-1,0,0,0,0),\\
(1,-1,0,0,0),\\
(0,1,-1,0,0),\\
(0,0,1,-1,0),\\
(0,0,0,1,-1),\\
(0,0,0,0,1)
\end{array}\right\}.}
$$
Every entry has <weight multiplicity> one.
The <two-index antisymmetric representation> is $\Lambda^2V=\mathbf{15}$, of dimension $\binom62=15$ and <highest weight> $w_1+w_2=(0,1,0,0,0)$. A <weight basis> is $e_i\wedge e_j$ for $i<j$, carrying <weight> $w_i+w_j$. The fifteen sums, all distinct and all of <weight multiplicity> one, are
$$
\boxed{
\begin{array}{c|rrrrr}
ij&a_1&a_2&a_3&a_4&a_5\\\hline
12&0&1&0&0&0\\
13&1&-1&1&0&0\\
14&1&0&-1&1&0\\
15&1&0&0&-1&1\\
16&1&0&0&0&-1\\
23&-1&0&1&0&0\\
24&-1&1&-1&1&0\\
25&-1&1&0&-1&1\\
26&-1&1&0&0&-1\\
34&0&-1&0&1&0\\
35&0&-1&1&-1&1\\
36&0&-1&1&0&-1\\
45&0&0&-1&0&1\\
46&0&0&-1&1&-1\\
56&0&0&0&-1&0
\end{array}.}
$$
The exclusion of $i=j$ comes from the <exterior square>, where $e_i\wedge e_i=0$.
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