= Solution
Choose the <maximal torus> of $SO(2n)$ consisting of rotations in $n$ <orthogonal> planes, and write $z_i=e^{i\theta_i}$. In the complexified standard <representation> $E=\mathbb C^{2n}$ the <torus> <eigenvalues> are $z_1,z_1^{-1},\ldots,z_n,z_n^{-1}$, with <weights> $e_i,-e_i$. For $n\geq2$ the complexified <Lie algebra> has the <Dn root system> $\{\pm e_i\pm e_j:i<j\}$. The <root-space decomposition> theorem splits this <Lie algebra> into its <Cartan subalgebra> and one-dimensional <root spaces>. The <Cartan subalgebra> contributes the zero <weight> with multiplicity $n$. Therefore its <character> is
$$
\boxed{\chi_{\mathrm{ad}}(z)=n+\sum_{i<j}\left(z_i z_j+\frac{z_i}{z_j}+\frac{z_j}{z_i}+\frac1{z_i z_j}\right).}
$$
Take <positive roots> $e_i-e_j,e_i+e_j$ for $i<j$, so $\rho=(n-1,n-2,\ldots,0)$. For $n>2$ the <root> $\theta=e_1+e_2$ is a <highest weight>: its <root vector> is killed by every <positive root> vector, since $\theta+\alpha$ is never a <root> for positive $\alpha$. To verify irreducibility without relying on an unstated simplicity theorem, apply complete reducibility and compare <dimensions>. For the <Dn root system> the <Weyl dimension formula> reads
$$
\dim V_\lambda=\prod_{i<j}\frac{(\lambda_i+\rho_i)^2-(\lambda_j+\rho_j)^2}{\rho_i^2-\rho_j^2}.
$$
For $\lambda=(1,1,0,\ldots,0)$ only pairs involving the first two indices change. With $s=0,\ldots,n-3$, cancellation gives
$$
\dim V_\theta
=\frac{2n-1}{2n-3}\prod_{s=0}^{n-3}\frac{n^2-s^2}{(n-2)^2-s^2}
=\frac{2n-1}{2n-3}\left(\frac{n(n-1)}2\right)\left(\frac{2(2n-3)}{n-1}\right)
=n(2n-1).
$$
This is exactly $\dim\mathfrak{so}_{2n}(\mathbb C)$, so the <irreducible> summand containing the <highest-weight vector> exhausts the <Adjoint representation>. Thus \b[for $n>2$ the complexified <Adjoint representation> is <irreducible>, with <highest weight> $e_1+e_2$].
For $n=2$, the <roots> split into the two independent systems $\{\pm(e_1+e_2)\}$ and $\{\pm(e_1-e_2)\}$. The <Adjoint representation> splits into two <irreducibles> of <dimension> three, with respective <highest weights> $(1,1)$ and $(1,-1)$. Equivalently these are the self-dual and anti-self-dual exterior two-forms in four <dimensions>. Their <characters> are
$$
1+z_1z_2+(z_1z_2)^{-1},\qquad
1+z_1/z_2+z_2/z_1.
$$
They add to the displayed adjoint <character>. This agrees with the <Chiral decomposition of the complexified so4 Lie algebra>.
For a direct <character> comparison, put $s(z)=\sum_i(z_i+z_i^{-1})$. The <eigenvalue> formula for an <exterior square> gives
$$
\chi_{\Lambda^2 E}(z)=\frac12\left(s(z)^2-\sum_i(z_i^2+z_i^{-2})\right)
=n+\sum_{i<j}\left(z_i z_j+z_i/z_j+z_j/z_i+(z_i z_j)^{-1}\right).
$$
This is precisely $\chi_{\mathrm{ad}}$. More intrinsically, let $B$ be the nondegenerate <symmetric bilinear form> preserved by $SO(2n)$ and define
$$
\Psi(u\wedge v)(x)=B(v,x)u-B(u,x)v.
$$
This endomorphism is skew with respect to $B$, so $\Psi$ maps $\Lambda^2E$ to $\mathfrak{so}(E,B)$. In an <orthonormal basis> the bivectors $e_i\wedge e_j$ map to a basis of skew <skew-symmetric matrices>, proving that it is an <isomorphism>. Since $B$ is invariant,
$$
\Psi(gu\wedge gv)=g\Psi(u\wedge v)g^{-1}.
$$
Thus \b[the <exterior square> and the complexified <Adjoint representation> are naturally isomorphic], explaining the <character> equality without a <weight> calculation.
The <symmetric square> is not <irreducible>. The <inverse metric> tensor $\Omega$ is a nonzero invariant vector, and contraction with $B$ splits off its trivial line:
$$
\operatorname{Sym}^2E=\mathbb C\Omega\oplus\operatorname{Sym}^2_0E.
$$
The contraction sends $\Omega$ to $2n$, so this is indeed a <direct sum>. A <highest-weight vector> $f_1\otimes f_1$, where $f_1$ has <torus> <weight> $e_1$, lies in the traceless part because $B(f_1,f_1)=0$, and has <highest weight> $2e_1$. The same <dimension> formula gives, for $n\geq2$,
$$
\dim V_{2e_1}=\prod_{s=0}^{n-2}\frac{(n+1)^2-s^2}{(n-1)^2-s^2}
=\left(\frac{n(n+1)}2\right)\left(\frac{2(2n-1)}n\right)
=(n+1)(2n-1).
$$
This equals $\dim\operatorname{Sym}^2E-1=n(2n+1)-1$. Complete reducibility therefore proves that the traceless part is <irreducible>, including the dimension-nine case for $SO(4)$. In particular,
$$
\boxed{\operatorname{Sym}^2(\mathbb C^{2n})\cong\mathbf1\oplus V_{2e_1}\quad(n\geq2),\quad\text{so it is reducible}.}
$$
If the circle case $n=1$ is included, the adjoint <character> is $1$ and the <symmetric square> instead has three one-dimensional <weights> $2,0,-2$.
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