= Solution
Start with the <tensor product of Lie algebra representations>, whose action is
$$
x\cdot(v\otimes w)=(xv)\otimes w+v\otimes(xw).
$$
Writing $T_x=\rho(x)\otimes I+I\otimes\rho(x)$, the two tensor factors commute, so $[T_x,T_y]=T_{[x,y]}$. This verifies the <Lie algebra representation> identity over every field.
The <exterior square> and <symmetric square> are the <quotient vector spaces>
$$
\bigwedge^2V=(V\otimes V)/\langle v\otimes v:v\in V\rangle,
\qquad
S^2V=(V\otimes V)/\langle v\otimes w-w\otimes v:v,w\in V\rangle.
$$
In the <exterior square>, expanding $(v+w)\otimes(v+w)$ shows that $v\wedge w=-w\wedge v$, including in <characteristic two>. Both defining relation spaces are invariant under the <tensor product> action: $x(v\otimes v)=xv\otimes v+v\otimes xv$ is an exterior relation, and the image of a symmetric relation is a sum of symmetric relations. Thus the quotient actions are well-defined and satisfy
$$
x(v\wedge w)=xv\wedge w+v\wedge xw,\qquad x(vw)=(xv)w+v(xw).
$$
For the printed <basis> $v_1,\ldots,v_n$, <bases> are $v_i\wedge v_j$ with $i<j$ and $v_iv_j$ with $i\le j$. Their dimensions are $n(n-1)/2$ and $n(n+1)/2$ respectively.
\b[If $2$ is invertible in $k$, $V\otimes V\cong\bigwedge^2V\oplus S^2V$ as representations.] Define the flip $\tau(v\otimes w)=w\otimes v$. It commutes with the <Lie algebra> action and satisfies $\tau^2=I$. Therefore
$$
P_+=\frac{I+\tau}{2},\qquad P_-=\frac{I-\tau}{2}
$$
are complementary invariant <linear projections>. The maps
$$
vw\longmapsto\frac{v\otimes w+w\otimes v}{2},\qquad
v\wedge w\longmapsto\frac{v\otimes w-w\otimes v}{2}
$$
identify $S^2V$ with $\operatorname{im}P_+$ and $\bigwedge^2V$ with $\operatorname{im}P_-$. Their inverses are the corresponding quotient maps restricted to these subspaces. This proves the assertion for every field of odd <characteristic>, and also for <characteristic zero>.
\b[Over every field, $S^2(V\oplus W)\cong S^2V\oplus S^2W\oplus(V\otimes W)$.] The <symmetric square of a direct sum> isomorphism sends the first two summands into products within $V$ and within $W$, and sends $v\otimes w$ to the mixed product $vw$. If $v_i$ and $w_j$ are <bases>, the monomial <basis> of $S^2(V\oplus W)$ is the disjoint union
$$
\{v_iv_j:i\le j\},\qquad\{w_iw_j:i\le j\},\qquad\{v_iw_j:\text{all }i,j\}.
$$
Thus the map is bijective, with no division by $2$ needed. The <Leibniz rule> for the action preserves each of these three summands and agrees with its usual <Lie algebra representation> action, proving equivariance.
For $\mathfrak g=\mathfrak{sl}(3)$, $V=\Gamma_{0,0}$ is trivial and $W=\Gamma_{2,1}$ has dimension $15$. Hence
$$
S^2(V\oplus W)\cong\Gamma_{0,0}\oplus\Gamma_{2,1}\oplus S^2\Gamma_{2,1}.
$$
It remains to find the <irreducible representations> in the <symmetric square of the sl3 representation of highest weight (2,1)>. We give the <formal character> calculation explicitly.
Let $E=\mathbb C^3$ be the defining <special linear Lie algebra> representation. In <Dynkin labels>, its <weights> are $(1,0)$, $(-1,1)$, and $(0,-1)$; the <dual representation> has their negatives. The equivariant contraction
$$
c:S^2E\otimes E^*\longrightarrow E,\qquad c(uv\otimes f)=f(u)v+f(v)u
$$
is surjective. Its <kernel> has dimension $18-3=15$. The tensor $e_1^2\otimes e_3^*$ is a <highest-weight vector> of <highest weight> $(2,1)$ in that <kernel>. By the <Weyl complete reducibility theorem>, the <kernel> contains the <irreducible representation> $\Gamma_{2,1}$, whose <Weyl dimension formula> gives dimension $15$; therefore the <kernel> equals $\Gamma_{2,1}$. This yields
$$
\chi_W=\chi_{S^2E}\chi_{E^*}-\chi_E.
$$
Multiplying the six <weights> of $S^2E$ by the three <weights> of $E^*$ and subtracting those of $E$ gives the following full <weight multiplicity> list:
$$
\begin{array}{c|l}
\text{multiplicity}&\text{weights in Dynkin labels}\\\hline
1&(2,1),(3,-1),(2,-2),(1,-3),(0,2),(-1,-2),(-2,3),(-2,0),(-3,2)\\
2&(1,0),(0,-1),(-1,1)
\end{array}
$$
The multiplicities sum to $9+2\cdot3=15$.
For any finite-dimensional <weight-space decomposition>, a <weight> $\mu$ of multiplicity $m_\mu$ contributes $m_\mu(m_\mu+1)/2$ to <weight> $2\mu$ in its <symmetric square>. Distinct <weights> $\mu,\nu$ contribute $m_\mu m_\nu$ to $\mu+\nu$. Equivalently,
$$
\chi_{S^2W}(z)=\frac{\chi_W(z)^2+\chi_W(z^2)}{2}.
$$
Applying this to the displayed list gives all dominant <weight multiplicities> in the second column below. The remaining columns are the <weight multiplicities> of the candidate <irreducible representations>:
$$
\begin{array}{c|r|rrrrr}
\text{weight}&S^2W&\Gamma_{4,2}&\Gamma_{3,1}&\Gamma_{0,4}&\Gamma_{1,2}&\Gamma_{2,0}\\\hline
(4,2)&1&1&0&0&0&0\\
(5,0)&1&1&0&0&0&0\\
(2,3)&1&1&0&0&0&0\\
(3,1)&3&2&1&0&0&0\\
(0,4)&2&1&0&1&0&0\\
(1,2)&5&2&1&1&1&0\\
(2,0)&8&3&2&1&1&1\\
(0,1)&9&3&2&1&2&1
\end{array}
$$
For an explicit way to compute each irreducible column, set $z_1z_2z_3=1$ and use the <Weyl character formula> in the form
$$
\chi_{a,b}(z_1,z_2,z_3)=
\frac{\det\begin{pmatrix}z_1^{a+b+2}&z_1^{b+1}&1\\z_2^{a+b+2}&z_2^{b+1}&1\\z_3^{a+b+2}&z_3^{b+1}&1\end{pmatrix}}
{\det\begin{pmatrix}z_1^2&z_1&1\\z_2^2&z_2&1\\z_3^2&z_3&1\end{pmatrix}}.
$$
A monomial $z_1^{n_1}z_2^{n_2}z_3^{n_3}$ has <Dynkin labels> $(n_1-n_2,n_2-n_3)$. Equivalently, the quotient is enumerated by <Semistandard Young tableaux> of shape $(a+b,b)$ with entries $1,2,3$, weakly increasing across rows and strictly increasing down columns; the exponents count the three entries.
The five irreducible columns sum to the $S^2W$ column. These are all its dominant <weights>, and all five candidate characters have no other dominant <weights>. Every <Weyl group> orbit meets the dominant chamber, and <weight multiplicities> are constant on <Weyl group> orbits. Thus the table proves equality of the full <formal characters>, and the <Weyl complete reducibility theorem> gives
$$
\boxed{S^2\Gamma_{2,1}\cong\Gamma_{4,2}\oplus\Gamma_{3,1}\oplus\Gamma_{0,4}\oplus\Gamma_{1,2}\oplus\Gamma_{2,0}.}
$$
The <Weyl dimension formula> checks the result:
$$
60+24+15+15+6=120=\frac{15\cdot16}{2}.
$$
Consequently the requested decomposition is
$$
\boxed{S^2(\Gamma_{0,0}\oplus\Gamma_{2,1})\cong\Gamma_{0,0}\oplus\Gamma_{2,1}\oplus\Gamma_{4,2}\oplus\Gamma_{3,1}\oplus\Gamma_{0,4}\oplus\Gamma_{1,2}\oplus\Gamma_{2,0}.}
$$
Its total dimension is $1+15+120=136=16\cdot17/2$.
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