You might want to first carry out a ($x$-dependent) unitary transformation of $A(x)$, such that all diagonal elements are zero.$^\ast$
Then $A(x)=BC(x)-C(x)B$ with
$$B=\begin{pmatrix}
1&0&0\\
0&2&0\\
0&0&3
\end{pmatrix},\;\;
C_{ij}(x)=\begin{cases}
0&\text{if}\;i=j\\
\frac{A_{ij}(x)}{B_{ii}-B_{jj}}&\text{if}\;\;i\neq j.
\end{cases}
$$
$^\ast$ This is always possible for a traceless $A$, see for example https://math.stackexchange.com/q/267192/87355
For a $3\times 3$ real matrix the unitary has the single-parameter form
$$U(x)=\begin{pmatrix}
\cos\alpha(x)&\sin\alpha(x)&0\\
-\sin\alpha(x)&\cos\alpha(x)&0\\
0&0&1
\end{pmatrix}.$$