= Solution
Define the <exterior square> over the arbitrary field $k$ by $\bigwedge^2V=(V\otimes V)/\operatorname{span}\{v\otimes v:v\in V\}$. Write the image of $u\otimes v$ as $u\wedge v$. Then $v\wedge v=0$ and $u\wedge v=-v\wedge u$; the basis is $v_i\wedge v_j$ with $i<j$, in <characteristic two> as well. The <exterior-power Lie algebra representation> is
$$
\boxed{x\cdot(u\wedge v)=(xu)\wedge v+u\wedge(xv).}
$$
The <tensor product of Lie algebra representations> descends to this quotient: $x\cdot(v\otimes v)=xv\otimes v+v\otimes xv$ is a linear combination of square tensors, namely $(v+xv)\otimes(v+xv)-v\otimes v-xv\otimes xv$. On the wedge basis its coefficients follow directly from the original representing matrices. Expanding both actions shows $[\rho_{\wedge^2}(x),\rho_{\wedge^2}(y)]=\rho_{\wedge^2}([x,y])$.
The inclusions of $V$ and $W$ into their <direct sum> define the map
$$
\bigwedge^2V\oplus\bigwedge^2W\oplus(V\otimes W)\longrightarrow\bigwedge^2(V\oplus W),\qquad(a,b,v\otimes w)\longmapsto a+b+(v,0)\wedge(0,w).
$$
A combined <basis> shows that this sends a basis to the pure-$V$, pure-$W$ and mixed wedge basis vectors, with no overlap and no omission. The defining action on each wedge proves equivariance. Hence the <exterior square of a direct sum> gives
$$
\boxed{\bigwedge^2(V\oplus W)\cong\bigwedge^2V\oplus\bigwedge^2W\oplus(V\otimes W).}
$$
No division by $2$ is involved, so the proof remains valid in <characteristic two>.
For the complex <special linear Lie algebra> $\mathfrak{sl}_3$, write $\Gamma_{a,b}$ for the irreducible <highest-weight representation> with <Dynkin labels> $(a,b)$, and $V=\Gamma_{1,0}$. First the <sl3 decomposition of the symmetric-square dual tensor product> is
$$
A:=(S^2V)\otimes V^*\cong\Gamma_{2,1}\oplus\Gamma_{1,0}.
$$
Indeed contraction $c((uv)\otimes f)=f(u)v+f(v)u$ is a surjective <Lie algebra representation homomorphism> onto $V$. Its 15-dimensional <kernel> contains the <highest-weight vector> $e_1^2\otimes e_3^*$ of weight $(2,1)$. The <Weyl dimension formula> gives $\dim\Gamma_{2,1}=15$, and the <Weyl complete reducibility theorem> identifies the kernel and splits the map. With $B=\Gamma_{2,1}$, the direct-sum identity reduces the requested calculation to $\bigwedge^2B$, $B\otimes V$ and $\bigwedge^2V$.
For completeness, these decompositions can be checked entirely by <formal characters>. Let $x_1x_2x_3=1$, $h_1=x_1+x_2+x_3$, $h_1^*=x_1^{-1}+x_2^{-1}+x_3^{-1}$ and $h_2=\sum_i x_i^2+\sum_{i<j}x_ix_j$. Then $\chi_B=h_2h_1^*-h_1$. The <Weyl character formula> takes the determinant form
$$
\chi_{a,b}(x)=\frac{\det\!\begin{pmatrix}x_1^{a+b+2}&x_1^{b+1}&1\\x_2^{a+b+2}&x_2^{b+1}&1\\x_3^{a+b+2}&x_3^{b+1}&1\end{pmatrix}}{\det\!\begin{pmatrix}x_1^2&x_1&1\\x_2^2&x_2&1\\x_3^2&x_3&1\end{pmatrix}}.
$$
Substitute $\chi_B$ into $\chi_{\wedge^2B}(x)=\tfrac12(\chi_B(x)^2-\chi_B(x_1^2,x_2^2,x_3^2))$ and collect the determinant characters. This gives the <exterior square of the sl3 representation of highest weight (2,1)> and the <sl3 highest-weight tensor rule>:
$$
\begin{aligned}
\bigwedge^2B&\cong\Gamma_{5,0}\oplus\Gamma_{2,3}\oplus\Gamma_{3,1}\oplus\Gamma_{1,2}\oplus\Gamma_{0,1},\\
B\otimes V&\cong\Gamma_{3,1}\oplus\Gamma_{1,2}\oplus\Gamma_{2,0},\qquad\bigwedge^2V\cong\Gamma_{0,1}.
\end{aligned}
$$
The first line has dimensions $21+42+24+15+3=105=\binom{15}{2}$; the tensor product has dimension $24+15+6=45$. Equivalently, enumerate weights with the <sl3 interlacing character formula> and subtract characters from the highest weight downwards. Finally
$$
\boxed{\bigwedge^2\!\bigl((S^2V)\otimes V^*\bigr)\cong\Gamma_{5,0}\oplus\Gamma_{2,3}\oplus2\Gamma_{3,1}\oplus2\Gamma_{1,2}\oplus\Gamma_{2,0}\oplus2\Gamma_{0,1}.}
$$
Its dimension is $21+42+48+30+6+6=153=\binom{18}{2}$. The exterior square here is taken of the entire 18-dimensional tensor product; taking it only of $S^2V$ would be a different representation.
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