Is this simple condition enough to show $f:\mathbb{C}^3\to\mathbb{C}^3$ is holomorphic? Let
$$
f:\mathbb{C}^3\to\mathbb{C}^3,\quad f(z)=(f_1(z),f_2(z),f_3(z))
$$
be a smooth map such that the three functions
$$
\begin{align}
z_2f_3(z)-z_3f_2(z)\\
z_3f_1(z)-z_1f_3(z)\\
z_1f_2(z)-z_2f_1(z)
\end{align}
$$
are holomorphic, where $z=(z_1,z_2,z_3)\in\mathbb{C}^3$. Is $f$ holomorphic? 
Note that these functions are obtained by cyclic permutations of $(1,2,3)$.
 A: No,it‘s not necessarily holomorphic.
We set \begin{align}
G_1(z)=z_2f_3(z)-z_3f_2(z)\\
G_2(z)=z_3f_1(z)-z_1f_3(z)\\
G_3(z)=z_1f_2(z)-z_2f_1(z)
\end{align}
Then,we have $\frac{\partial G_i}{\partial\bar z_k}=0$,for $i,k=1,2,3$.
Hence,we have 
$$ \left[ {\begin{array}{*{20}{c}}
0 & z_3 &-z_2&0&0&0&0&0&0\\
z_2 & -z_1&0&0&0&0&0&0&0\\
z_3&0&-z_1&0&0&0&0&0&0\\
0&0&0&0 & z_3 &-z_2&0&0&0\\
0&0&0&z_2 & -z_1&0&0&0&0\\
0&0&0&z_3&0&-z_1&0&0&0\\
0&0&0&0&0&0&0 & z_3 &-z_2\\
0&0&0&0&0&0&z_2 & -z_1&0\\
0&0&0&0&0&0&z_3&0&-z_1\\
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
\frac {\partial f_1}{\partial\bar z_1}\\
\frac {\partial f_2}{\partial\bar z_1}\\
\frac {\partial f_3}{\partial\bar z_1}\\
\frac {\partial f_1}{\partial\bar z_2}\\
\frac {\partial f_2}{\partial\bar z_2}\\
\frac {\partial f_3}{\partial\bar z_2}\\
\frac {\partial f_1}{\partial\bar z_3}\\
\frac {\partial f_2}{\partial\bar z_3}\\
\frac {\partial f_3}{\partial\bar z_3}\\
\end{array}} \right]=0$$
Notice that the determinant of the matrix equal to zero,so we can't show
$\frac{\partial f_i}{\partial\bar z_k}=0$,for $i,k=1,2,3$.
As Conrad said ,we can take $f_k(z)=z_k(\bar z_1+\bar z_2+\bar z_3)$ as a counterexample.
