= Solution
In the <defining representation> $V=\mathbb C^{2n}$, the basis vectors $v_i$ and $v_{\bar i}$ have <weights> $\varepsilon_i$ and $-\varepsilon_i$, respectively, each with multiplicity one. Thus its <formal character> is
$$
\boxed{\operatorname{ch}V=\sum_{i=1}^n(e^{\varepsilon_i}+e^{-\varepsilon_i}),\qquad\text{highest weight }\varepsilon_1.}
$$
The vector $v_1$ is killed by every positive <root space>, since all positive-root matrices are upper triangular, and is a <highest-weight vector>. Any nonzero invariant subspace splits into weight spaces and therefore contains one of the natural weight vectors, since those weight spaces are one-dimensional. The root matrices connect that vector to all the others, so this representation is <irreducible> for $n\geq2$. In terms of the <fundamental weights of Dn>, its highest weight is $\omega_1$ for $n\geq3$ and $\omega_1+\omega_2$ for $n=2$.
The flip of the two tensor factors is equivariant, hence
$$
V\otimes V=\operatorname{Sym}^2V\oplus\Lambda^2V.
$$
Use the invariant <symmetric bilinear form> $Q(u,v)=u^TJv$. Contraction $C(u\otimes v)=Q(u,v)$ is an equivariant map to the <trivial Lie algebra representation>. Its inverse-form tensor
$$
\tau=\sum_{i=1}^n(v_i\otimes v_{\bar i}+v_{\bar i}\otimes v_i)
$$
is invariant and satisfies $C(\tau)=2n$. Consequently
$$
\operatorname{Sym}^2V=\mathbb C\tau\oplus\operatorname{Sym}^2_0V,\qquad\operatorname{Sym}^2_0V=\ker(C|_{\operatorname{Sym}^2V}).
$$
The traceless summand contains the <highest-weight vector> $v_1\otimes v_1$ of <weight> $2\varepsilon_1$. The <Weyl complete reducibility theorem> states that every finite-dimensional complex <semisimple Lie algebra> representation is a direct sum of irreducible representations, so the submodule generated by this highest vector is $L(2\varepsilon_1)$. The <Weyl dimension formula> verifies that it is the entire traceless space: only positive roots involving $\varepsilon_1$ change their factors, and
$$
\dim L(2\varepsilon_1)=\prod_{j=2}^n\frac{j+1}{j-1}\frac{2n-j+1}{2n-j-1}=\frac{n(n+1)}2\frac{2(2n-1)}n=(n+1)(2n-1).
$$
This equals $\dim\operatorname{Sym}^2V-1=n(2n+1)-1$. Thus the <traceless symmetric square of the defining even orthogonal representation> is irreducible, including $n=2$.
For the other summand, the <exterior square of the defining orthogonal representation> has the equivariant identification
$$
u\wedge v\longmapsto\bigl[w\longmapsto Q(v,w)u-Q(u,w)v\bigr]\in\mathfrak{so}(V,Q).
$$
The image is skew relative to $Q$, and in a basis making $Q$ the identity the images are the independent elementary skew matrices. The map is therefore an isomorphism onto the <Adjoint representation of a Lie algebra>. For $n\geq3$, the connected <Dn Dynkin diagram> gives a <simple Lie algebra> (with $D_3=A_3$); its adjoint module is irreducible because its invariant subspaces are exactly its ideals. Its highest weight is $\varepsilon_1+\varepsilon_2$, and $v_1\wedge v_2$ is a highest vector. The <tensor-square decomposition of the defining even orthogonal representation> is consequently
$$
\boxed{V\otimes V=L(2\varepsilon_1)\oplus L(\varepsilon_1+\varepsilon_2)\oplus L(0)\qquad(n\geq3).}
$$
The three dimensions are $(n+1)(2n-1)$, $n(2n-1)$ and one, adding to $4n^2$. The highest weights are $2\omega_1$, $\omega_2$ and zero for $n\geq4$; for $n=3$ the middle weight is $\omega_2+\omega_3$.
For $n=2$, the <Chiral decomposition of the complexified so4 Lie algebra> splits the adjoint module into its two three-dimensional ideals. Thus
$$
\boxed{V\otimes V=L(2\varepsilon_1)\oplus L(\varepsilon_1-\varepsilon_2)\oplus L(\varepsilon_1+\varepsilon_2)\oplus L(0),\quad\dim=9+3+3+1.}
$$
The weights are $2\omega_1+2\omega_2$, $2\omega_1$, $2\omega_2$ and zero. Equivalently, labelling highest weights by the two <sl2 Lie algebra> factors, this is $(2,2)\oplus(2,0)\oplus(0,2)\oplus(0,0)$. If $n=1$ is allowed, the abelian algebra has $V=L(\varepsilon_1)\oplus L(-\varepsilon_1)$ and $V\otimes V=L(2\varepsilon_1)\oplus2L(0)\oplus L(-2\varepsilon_1)$; there is then no single highest weight for an irreducible natural module.
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