= Solution
Let $x_1,x_2,x_3$ be the coordinate basis of the <defining representation> $W$ of $\mathfrak{sl}_3$. Write $s_{ij}=x_ix_j$ for $1\leq i\leq j\leq3$. The six $s_{ij}$ form a basis of the <symmetric power> $\operatorname{Sym}^2W$, with weight $L_i+L_j$. A basis for the <tensor product of Lie algebra representations> $T$ consists of all 36 ordered tensors
$$
\boxed{s_{ij}\otimes s_{k\ell}\quad(1\leq i\leq j\leq3,\ 1\leq k\leq\ell\leq3).}
$$
The order of the two factors matters; there is no quotient identifying their interchange.
Each <weight of a representation> has the form $k_1L_1+k_2L_2+k_3L_3$, where the nonnegative <integers> satisfy $k_1+k_2+k_3=4$. Its <weight multiplicity> is the number of ways to split this triple into two nonnegative triples, each summing to two. The possible patterns, including every permutation, are
$$
\begin{array}{c|c|c}
(k_1,k_2,k_3)\text{ up to permutation}&\text{number of weights}&\text{multiplicity}\\\hline
(4,0,0)&3&1\\
(3,1,0)&6&2\\
(2,2,0)&3&3\\
(2,1,1)&3&4
\end{array}
$$
For example, $(2,2,0)$ splits as $(2,0,0)+(0,2,0)$, the reverse split, or $(1,1,0)+(1,1,0)$, giving multiplicity three. For $(2,1,1)$, the four splits pair $(2,0,0)$ with $(0,1,1)$ and $(1,1,0)$ with $(1,0,1)$, each in both orders. The multiplicities sum to $3+12+9+12=36$, verifying the dimension.
The following <weight diagram> uses <Dynkin labels> $(p,q)=(k_1-k_2,k_2-k_3)$; the circle numbers give multiplicities and orange points mark the <Closed dominant Weyl chamber>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-3-sl3-weights.png]
{title=The fifteen weights and multiplicities of Sym2 C3 tensor Sym2 C3}
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