[proofplan]
We work in the chosen holomorphic coordinate frame and write the metric as the positive definite Hermitian matrix $H=(h_{i\bar j})$. The local Chern curvature convention gives $\Theta^C=\bar\partial(H^{-1}\partial H)$, so the problem reduces to computing the trace of the matrix-valued form $H^{-1}\partial H$. The determinant identity $\operatorname{tr}(H^{-1}\partial H)=\partial\log\det H$ follows from the differential of the determinant, and the sign comes from the anticommutation relation $\bar\partial\partial=-\partial\bar\partial$ on scalar functions.
[/proofplan]
[step:Write the Chern curvature trace in the coordinate frame]
On $U$, let
\begin{align*}
A=H^{-1}\partial H
\end{align*}
denote the matrix-valued $(1,0)$-form of the Chern connection in the holomorphic coordinate frame $(e_1,\dots,e_n)$. By the stated curvature convention,
\begin{align*}
\Theta^C=\bar\partial A=\bar\partial(H^{-1}\partial H).
\end{align*}
The trace map is complex-linear on matrix entries and commutes with applying $\bar\partial$ entrywise. Hence
\begin{align*}
\operatorname{tr}\Theta^C=\operatorname{tr}\bar\partial(H^{-1}\partial H)=\bar\partial\operatorname{tr}(H^{-1}\partial H).
\end{align*}
[/step]
[step:Compute the trace of $H^{-1}\partial H$ by differentiating the determinant]
Let $G:U\to GL_n(\mathbb{C})$ denote the inverse matrix-valued function $G=H^{-1}$. Thus the entries $g_{\bar j i}$ of $G$ are defined by
\begin{align*}
\sum_{j=1}^n h_{i\bar j}g_{\bar j k}=\delta_{ik}
\end{align*}
for all $1\le i,k\le n$.
We claim that
\begin{align*}
\operatorname{tr}(H^{-1}\partial H)=\partial\log\det H.
\end{align*}
Indeed, since $H(x)$ is positive definite Hermitian for every $x\in U$, $\det H(x)>0$, so $\log\det H$ is a well-defined smooth real-valued function on $U$. The differential of the determinant at an invertible matrix $M\in GL_n(\mathbb{C})$ in the direction of a matrix $B\in M_n(\mathbb{C})$ is
\begin{align*}
d(\det)_M(B)=\det(M)\operatorname{tr}(M^{-1}B).
\end{align*}
Applying this pointwise to the smooth matrix-valued function $H$ gives
\begin{align*}
\partial(\det H)=(\det H)\operatorname{tr}(H^{-1}\partial H).
\end{align*}
Dividing by the positive function $\det H$ yields
\begin{align*}
\partial\log\det H=\frac{\partial(\det H)}{\det H}=\operatorname{tr}(H^{-1}\partial H).
\end{align*}
[guided]
The only non-formal identity in the proof is the determinant differentiation formula, so we verify exactly how it is being used. The map $H:U\to GL_n(\mathbb{C})$ is smooth because the metric coefficients $h_{i\bar j}$ are smooth in the holomorphic coordinate chart, and $H(x)$ is invertible for each $x\in U$ because $h_x$ is positive definite. Therefore $\det H:U\to(0,\infty)$ is a smooth positive function, and $\log\det H$ is well-defined.
For an invertible complex matrix $M\in GL_n(\mathbb{C})$ and a complex matrix $B\in M_n(\mathbb{C})$, the derivative of the determinant in the direction $B$ is
\begin{align*}
d(\det)_M(B)=\det(M)\operatorname{tr}(M^{-1}B).
\end{align*}
Here the role of $M$ is played by the pointwise matrix $H(x)$, and the role of $B$ is played by the matrix of $(1,0)$-forms $\partial H=(\partial h_{i\bar j})_{i,j=1}^n$. Applying the formula entrywise gives the $(1,0)$-form identity
\begin{align*}
\partial(\det H)=(\det H)\operatorname{tr}(H^{-1}\partial H).
\end{align*}
Since $\det H$ is positive and hence nowhere zero, we may divide by it. The usual chain rule for the smooth function $\log:(0,\infty)\to\mathbb{R}$ gives
\begin{align*}
\partial\log\det H=\frac{\partial(\det H)}{\det H}.
\end{align*}
Combining the two displayed identities yields
\begin{align*}
\partial\log\det H=\operatorname{tr}(H^{-1}\partial H).
\end{align*}
This is the matrix identity that converts the curvature trace into a scalar Dolbeault expression.
[/guided]
[/step]
[step:Use Dolbeault anticommutation to obtain the sign]
Substituting the determinant identity into the curvature trace computation gives
\begin{align*}
\operatorname{tr}\Theta^C=\bar\partial\partial\log\det H.
\end{align*}
For a smooth scalar function $f:U\to\mathbb{C}$, the Dolbeault operators satisfy
\begin{align*}
\bar\partial\partial f=-\partial\bar\partial f.
\end{align*}
Applying this with $f=\log\det H$ gives
\begin{align*}
\operatorname{tr}\Theta^C=-\partial\bar\partial\log\det H.
\end{align*}
[/step]
[step:Multiply by the Chern-Ricci convention]
By the stated convention for the Chern-Ricci form,
\begin{align*}
\operatorname{Ric}^C(h)=i\operatorname{tr}\Theta^C.
\end{align*}
Using the trace identity just obtained,
\begin{align*}
\operatorname{Ric}^C(h)=i\left(-\partial\bar\partial\log\det H\right)=-i\partial\bar\partial\log\det H.
\end{align*}
Since $H=(h_{i\bar j})$, this is precisely
\begin{align*}
\operatorname{Ric}^C(h)=-i\partial\bar\partial\log\det(h_{i\bar j})
\end{align*}
on the coordinate chart $U$.
[/step]
