Let $(M, g, J, \omega)$ be a Kähler manifold. That is, $(M, J)$ is a complex manifold, $g$ is a Hermitian metric on $M$ and $$\omega (X, Y) = g(JX, Y)$$ is a closed two form.
As a Riemannian manifold $(M, g)$, for each $x\in M$, one can find a geodesic normal coordinates around each $x$. In the case of Kähler metric, one actually have more:
Proposition: (Complex Normal Coordinates on Kähler manifolds) For each $x\in M$, there is a local holomorphic coordinates around $x$ so that the metric $g = g_{i\bar j}$ satisfies $$g_{i\bar j}(x) = \delta_{ij}, \ \ d g_{i\bar j} (x) = 0, \ \ \ \frac{\partial^2 g_{i\bar j}}{\partial z_k \partial z_l} (x) = 0.$$
While the first two conditions are similar to what we have for geodesic normal coordinates in Riemannian geometry, there is no corresponding analogue for the last condition. Also, even in a Kähler manifold, the geodesic normal coordinates might not be holomorphic.
I am looking for a proof of this proposition.
The result is local, we may assume that $p =0 \in U$, where $U \subset \mathbb C^n$ is an open set. By an complex linear transformation, we may also assume $h_{i\bar j}(0) = \delta_{ij}$. Consider the holomorphic mapping $\phi : B_\epsilon \to U$, given by $$z_i=\phi (w)_i = w_i + A^i_{mn} w_mw_n + B^i_{pqr} w_pw_qw_r,$$
here $A^i_{mn}, B^i_{pqr}$ are constants to be chosen, and repeated indices means sum over. We also assume that $A^i_{mn}, B^i_{pqr}$ are fully symmetric with respect to the lower indices. Since $d\phi_0 = Id$, $\phi$ is an biholomorphism onto its image (restricting to smaller sets when necessary). Let $g = \phi^* h$ be the pullback metric. Then using the new cordinates $(w_1, \cdots, w_n)$,
\begin{align}\tag{1} g_{\alpha \bar \beta} &= \frac{\partial z_i}{\partial w^\alpha}\overline{\frac{\partial z_j}{\partial w^\beta}}h_{i\bar j} \end{align} and $$\tag{2} \frac{\partial z_i}{\partial w_\alpha} = \delta_{i\alpha} + 2A^i_{\alpha m} w_m + 3B^i_{\alpha pq } w_pw_q$$ Then we have \begin{align}g_{\alpha\bar\beta} &= h_{\alpha\bar\beta} +2A^i_{\alpha m} h_{i\bar\beta}w_n + 2\overline{A^j_{\beta n}} h_{\alpha \bar j} \bar w_n \\ &\ \ \ + 3B^i_{\alpha pq } w_pw_q h_{i\bar\beta} + 3\overline{B^j_{\beta rs} } \bar w_r \bar w_s h_{\alpha \bar j} + 4 A^i_{\alpha m} \overline{A^j_{\beta n} } h_{i\bar j}w_m \bar w_n\\ &\ \ \ + O(|w|^3). \end{align} From here it's obvious that $g_{\alpha\bar\beta}(0) = \delta_{\alpha\beta}$. Also, we have \begin{align} \partial_\gamma g_{\alpha\bar\beta} (0) &= \partial_\gamma h_{\alpha\bar\beta} (0) + 2A^\beta_{\alpha\gamma}, \\ \bar\partial_\gamma g_{\alpha\bar\beta} (0) &= \bar\partial_\gamma h_{\alpha\bar\beta} (0) + 2\overline{A^\alpha_{\beta\gamma}}, \\ \partial_\eta \partial _\gamma g_{\alpha\bar\beta} (0) &= \partial_\eta \partial _\gamma h_{\alpha\bar\beta} (0) + 6B^\beta_{\alpha\gamma\eta}. \end{align}
Now we choose $A^\beta_{\alpha\gamma} = -\frac 12 \partial_\gamma h_{\alpha\bar\beta}$. First of all, $A$ so chosen is really symmetric in the lower indice, since $$\tag{3}\partial_\gamma h_{\alpha\bar\beta} =\partial_\alpha h_{\gamma \bar\beta}$$ when $h$ is a Kähler metric (see here). Next, since $h$ is Hermitian, $$\bar \partial_\gamma h_{\alpha\bar\beta} (0) = \overline{\partial_\gamma \overline{h_{\alpha\bar\beta}(0)}} = \overline{\partial_\gamma h_{\beta\bar\alpha}(0)} = -\frac{1}{2} \overline{A^\alpha_{\beta\gamma}}, $$ where in the last equality we used the definition of $A$. Thus we have $dg_{i\bar j}(0) = 0$.
Lastly, we choose $B^\beta_{\alpha\gamma\delta} = - \partial_\eta \partial _\gamma h_{\alpha\bar\eta} (0)$. Again by (3), $\partial_\eta \partial _\gamma h_{\alpha\bar\eta}$ is symmetric in $\alpha, \gamma, \eta$. Thus $B$ is again well defined, and this finishes the proof.
In this book, they state the following proposition:
They also suggest a reference, p.286, claiming that there's an elegant proof.