= Solution
Fix the convention that the <Hermitian metric> is conjugate-linear in its first argument. In a <holomorphic local frame> $e=(e_1,\ldots,e_r)$, let $H_{ij}=h(e_i,e_j)$, so $h(ev,ew)=v^\dagger Hw$. Write $De=e\theta$, or equivalently $D(ev)=e(dv+\theta v)$.
Compatibility with the <holomorphic vector bundle> structure means $D^{0,1}=\bar\partial_E$. Since the frame is holomorphic, this forces $\theta^{0,1}=0$. <Metric compatibility> means
$$
dh(s,t)=h(Ds,t)+h(s,Dt),
$$
or, in the chosen frame, $dH=\theta^\dagger H+H\theta$. Conjugate transpose here also conjugates the one-form factors. Taking the $(1,0)$ component gives $\partial H=H\theta$, and hence
$$
\boxed{\theta=H^{-1}\partial H.}
$$
This proves uniqueness and suggests existence. The proposed $\theta$ has type $(1,0)$, and its conjugate transpose obeys $\theta^\dagger H=\bar\partial H$, since $H=H^\dagger$. Therefore $dH=\theta^\dagger H+H\theta$, proving <metric compatibility> as well.
Under a change of <holomorphic local frame> $e'=eG$, we have $H'=G^\dagger HG$ and $\partial G^\dagger=0$. A direct computation yields
$$
(H')^{-1}\partial H'=G^{-1}\theta G+G^{-1}\partial G.
$$
Because $G$ is holomorphic, $\partial G=dG$; this is exactly the transformation rule for a <connection one-form>. The locally defined connections therefore glue to a global <connection on a vector bundle>. \b[The Chern connection exists uniquely], and its <local formula for the Chern connection on a vector bundle> is
$$
\boxed{D=d+H^{-1}\partial H,\qquad D^{0,1}=\bar\partial_E.}
$$
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