= Solution
Use the convention that a <Hermitian metric> $h$ is complex-linear in its first argument and conjugate-linear in its second. It is a smoothly varying <positive-definite> <Hermitian form> on each fibre. A <connection on a vector bundle> is a complex-linear operator $D:\Gamma(E)\to\Omega^1(X,E)$ satisfying $D(as)=da\otimes s+aDs$ for smooth complex functions $a$. Metric compatibility means, for every real <vector field> $V$,
$$
\boxed{Vh(s,t)=h(D_Vs,t)+h(s,D_Vt).}
$$
This is the <metric-compatible connection> condition with the sesquilinear convention fixed.
Apply the smooth <Gram-Schmidt process> to a local frame to obtain an $h$-orthonormal smooth frame $s_1,\ldots,s_r$. Positivity guarantees that all normalization denominators are nonzero and depend smoothly on the base point. Write $Ds_j=\sum_i A_{ij}s_i$. Differentiating $h(s_i,s_j)=\delta_{ij}$ and using compatibility gives
$$
A_{ji}(V)+\overline{A_{ij}(V)}=0
$$
for every real $V$. Thus $\boxed{A(V)^\dagger=-A(V)}$: the connection matrix is skew-Hermitian. This <smooth unitary frame for a Hermitian connection> is generally not holomorphic; the requested local-frame assertion requires only a smooth frame.
The <holomorphic dual vector bundle> $E^*$ is obtained by dualizing fibres and using transition matrices $g_{ij}^{-T}$ when $E$ has transitions $g_{ij}$. Their entries are holomorphic because matrix inversion is holomorphic on $GL_r(\mathbb C)$. Their cocycle property follows from preservation of the fibrewise evaluation pairing, defining the natural holomorphic bundle with fibre $(E_x)^*$.
The <conjugate vector bundle> $\overline E$ has the same underlying real fibres but opposite scalar action: $\lambda\cdot\overline v=\overline{\overline\lambda v}$. Its smooth transition matrices are $\overline{g_{ij}}$; they need not be holomorphic on $X$. The conjugate bundle is used here as a smooth complex bundle, whereas $E^*$ is holomorphic.
Define the smooth dual tensor $H$ by
$$
H(s\otimes\overline t)=h(s,t).
$$
It is complex-linear in each tensor factor, because conjugating the second bundle converts the conjugate-linearity of $h$ into linearity. Thus $H\in\Gamma((E\otimes\overline E)^*)$.
The <conjugate connection> is defined on real <vector fields> by $\overline D_V(\overline s)=\overline{D_Vs}$ and extended complex-linearly on the conjugate bundle. The <tensor product connection> is
$$
D_\otimes(s\otimes\overline t)=Ds\otimes\overline t+s\otimes\overline{Dt}.
$$
Its <dual connection> $D_0$ is uniquely characterized by
$$
(D_{0,V}H)(u)=V[H(u)]-H(D_{\otimes,V}u).
$$
The Leibniz rule makes this a genuine connection on $(E\otimes\overline E)^*$. Evaluate it on the decomposable tensor $u=s\otimes\overline t$:
$$
\boxed{(D_{0,V}H)(s\otimes\overline t)
=Vh(s,t)-h(D_Vs,t)-h(s,D_Vt).}
$$
Decomposable tensors span every fibre. Hence this tensor-valued one-form vanishes precisely when the compatibility identity holds for every $V,s,t$:
$$
\boxed{D\text{ is compatible with }h\ \Longleftrightarrow\ D_0H=0.}
$$
This is <metric compatibility as parallelism of a Hermitian tensor>.
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