= Solution
Write $x=(x_1,x_2,x_3)^T$, $v=(x_4,x_5,x_6)^T$ and $z=x+iv$. Here products of differentials in a <metric tensor> are symmetric products, unlike the <wedge products> used later. Then
$$
\boxed{g=\sum_{k=1}^3 dz_k\,d\bar z_k
=\frac12\sum_{k=1}^3(dz_k\otimes d\bar z_k+d\bar z_k\otimes dz_k).}
$$
Indeed $dz_k\,d\bar z_k=dx_k^2+dx_{k+3}^2$. A transformation $z'=Uz$ preserves this <metric> whenever $U^\dagger U=I$, because
$$
\sum_k d\bar z'_k\,dz'_k=d\bar z^T U^\dagger U\,dz=d\bar z^Tdz.
$$
Thus the entire <unitary group> $U(3)$ preserves it, including its <special unitary group> $SU(3)$.
To obtain a real <matrix> embedding, write $U=A+iB$ with $A,B$ real. Then
$$
\boxed{\rho:SU(3)\hookrightarrow SO(6),\qquad
\rho(U)=\begin{pmatrix}A&-B\\B&A\end{pmatrix}.}
$$
The <unitarity> equations imply $A^TA+B^TB=I$ and $A^TB=B^TA$, hence $\rho(U)^T\rho(U)=I$. Multiplication of the block <matrices> gives $\rho(UV)=\rho(U)\rho(V)$, and the blocks recover $U$, proving injectivity. Finally
$$
\det_{\mathbb R}\rho(U)=\det_{\mathbb C}U\,\det_{\mathbb C}\bar U=|\det_{\mathbb C}U|^2=1.
$$
For this last identity, over the <complex numbers> the change of <basis> from $(x,v)$ to $(z,\bar z)$ conjugates $\rho(U)$ to $\operatorname{diag}(U,\bar U)$. This proves that the image lies in the <special orthogonal group>.
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