= Solution
Define an equivariant contraction $C:V^*\otimes\operatorname{Sym}^2V\to V$ by
$$
C(\varphi\otimes uv)=\varphi(u)v+\varphi(v)u.
$$
This uses the natural dual pairing, so it commutes with the <Lie algebra representation>. Also define $s(v)=\sum_i e_i^*\otimes e_iv$. The identity tensor $\sum_i e_i^*\otimes e_i$ is invariant, making $s$ equivariant. Direct contraction gives $C(s(v))=3v+v=4v$. Thus $s$ is injective and $C$ is surjective, and
$$
w=\left(w-\frac14s(Cw)\right)+\frac14s(Cw)
$$
is a direct decomposition into $\ker C$ and $s(V)$. In particular $\dim\ker C=18-3=15$ and $s(V)$ is a defining irreducible module of <highest weight> $\omega_1$.
For the usual upper-triangular choice of positive roots, $T_3^{11}=z^*\otimes x^2$ lies in the kernel and is killed by both simple raising operators $E_{12},E_{23}$. Its highest weight is $2\epsilon_1-\epsilon_3=2\omega_1+\omega_2$, with <Dynkin labels> $(2,1)$. The <highest-weight classification> and complete reducibility give an irreducible copy $L(2,1)$ generated by this vector. The <Weyl dimension formula> for $\mathfrak{sl}_3$ is
$$
\dim L(p,q)=\frac{(p+1)(q+1)(p+q+2)}2,
$$
so $\dim L(2,1)=15$. It fills the entire kernel. Therefore the <sl3 decomposition of the symmetric-square dual tensor product>, with its tensor factors exchanged, is
$$
\boxed{W=\ker C\oplus s(V)\cong L(2,1)\oplus L(1,0),\qquad 18=15+3.}
$$
The weight diagrams below record both summands. Their labels use $K_{ij}=2T_j^{ij}-T_i^{ii}$ for $j\ne i$ and $S_i=\sum_jT_j^{ij}$; these are the weight bases detailed in part (iii).
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-1-sl3-summands.png]
{title=Annotated weight bases and highest weights of the fifteen-dimensional and defining sl3 summands}
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