= Solution
Let $e_1=x,e_2=y,e_3=z$, and let $\epsilon_i$ be the <weight of a representation> defined by $\epsilon_i(\operatorname{diag}(h_1,h_2,h_3))=h_i$. Since the trace is zero, $\epsilon_1+\epsilon_2+\epsilon_3=0$. The <dual representation> gives weight $-\epsilon_i$ to $e_i^*$. Put
$$
T_i^{jk}=e_i^*\otimes e_je_k=T_i^{kj}.
$$
Taking $j\le k$ gives an eighteen-vector <tensor product> basis. Its weight is $\epsilon_j+\epsilon_k-\epsilon_i$.
For explicit coordinates use the usual simple coroots $H_1=\operatorname{diag}(1,-1,0)$ and $H_2=\operatorname{diag}(0,1,-1)$, and write each weight as $(\mu(H_1),\mu(H_2))$. Then $\epsilon_1=(1,0)$, $\epsilon_2=(-1,1)$ and $\epsilon_3=(0,-1)$. Collecting equal weights gives every <weight space>:
$$
\begin{array}{c|c}
\text{weight coordinates}&\text{basis}\\\hline
(2,1)&T_3^{11}\\
(3,-1)&T_2^{11}\\
(-2,3)&T_3^{22}\\
(-3,2)&T_1^{22}\\
(1,-3)&T_2^{33}\\
(-1,-2)&T_1^{33}\\
(0,2)&T_3^{12}\\
(2,-2)&T_2^{13}\\
(-2,0)&T_1^{23}\\
(1,0)&T_1^{11},\ T_2^{12},\ T_3^{13}\\
(-1,1)&T_1^{12},\ T_2^{22},\ T_3^{23}\\
(0,-1)&T_1^{13},\ T_2^{23},\ T_3^{33}
\end{array}
$$
The first six weights are $2\epsilon_j-\epsilon_i$ with $i\ne j$; the next three are $-2\epsilon_i$; the last three are the defining weights $\epsilon_i$. Thus \b[there are nine multiplicity-one weights and three multiplicity-three weights], giving total dimension $9+9=18$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-1-sl3-weights.png]
{title=Weight diagram and tensor bases of the dual defining sl3 module tensor its symmetric square}
The planar coordinates are $a\omega_1+b\omega_2$ with $\omega_1=(1,0)$ and $\omega_2=(1/2,\sqrt3/2)$, so the <weight diagram> uses an equilateral root geometry rather than orthogonal Dynkin-coordinate axes. Every point is labelled with its complete weight-space basis.
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