Let $(M,J)$ be a complex manifold and $g$ be a Riemannian metric satisfying $g(J \cdot, J \cdot) = g(\cdot, \cdot)$. Let $\xi$ be a vector field and $\eta$ be the $1$-form defined by $\eta(X) := g(\xi, X)$.
Is it true that if $\bar\partial \eta = 0$ then $\xi$ is a holomorphic vector field, i.e. $L_\xi J = 0$? If not in general, what about if $M$ is compact Kähler? What about the converse?
About the converse, if $\xi$ is holomorphic then $d\eta$ is a $(1,1)$-form; if $M$ is Kähler, by the global $\partial \bar\partial$-lemma we can write $d\eta = dd^c u$ for some function $u$, so $\eta - d^c u$ is closed, so by Hodge decomposition $\eta - d^c u = df + \eta_0$ where $\eta_0$ is $\Delta$-harmonic, so $\Delta^\partial$-harmonic (because we're Kähler), so, since $M$ is compact $\partial \eta_0 = 0$. So $$\partial \eta = \partial d^c u + \partial(\partial + \bar\partial)f = i\partial (\bar\partial - \partial)u + \partial\bar\partial f =\partial\bar\partial (iu-f) $$ and on the other hand $d\eta = dd^c u$ because $d\eta_0 = 0$ and so $d \eta - \partial\eta = \bar\partial\eta = \partial\bar\partial f$. I don't know if there is any conclusion that can be drawn from here.
For the initial question, I don't know what would be an approach in the first place.
The form $\eta$ is holomorphic if and only if $\eta(X)$ is a holomorphic function for every holomorphic tangent field $X$. If $X$ is such a field we have $$ \bar\partial \eta(X) = \bar\partial g(X, \overline \xi) = g(X, \overline{D' \xi}), $$ where $D'$ is the $(1,0)$-part of the Chern connection of $g$. (I've assumed here that $g$ is linear in its first variable and antilinear in the second, so that $\eta$ is linear.) It follows that $\bar\partial \eta = 0$ if and only if $D'\xi = 0$. I don't think it necessarily follows that $\bar\partial \xi = 0$.
Suppose in fact that $M$ admits a flat Chern connection $D$ and let $\xi$ be a parallel holomorphic tangent field. If $f$ is a nonconstant holomorphic function, then $$ D'(\overline f \cdot \xi) = \partial \overline f \otimes \xi + \overline f D' \xi = 0 $$ but $\bar\partial(\overline f \xi) = \overline{\partial f} \otimes \xi \not= 0$, and $\overline f \xi$ should make for a holomorphic form $\eta$.