= Solution
The <Chern connection> of a <Hermitian metric on a holomorphic vector bundle> is the unique <connection on a vector bundle> compatible with that metric and whose $(0,1)$ part is the bundle's <Dolbeault operator> $\bar\partial_E$. Fix a <holomorphic local frame> and column coefficients for sections. Write the metric as $h(s,t)=\bar s^T Ht$, conjugate-linear in the first argument, and write the connection as $Ds=ds+A s$. The condition $D^{0,1}=\bar\partial_E$ forces $A$ to have type $(1,0)$. <Metric compatibility> requires
$$
dH=A^\dagger H+HA,
$$
where the dagger conjugates the differential-form coefficients as well as transposing the matrix. Taking the $(1,0)$ part gives \b[$\boxed{A=H^{-1}\partial H}$]. Its conjugate-transpose supplies the $(0,1)$ metric equation because $H$ is Hermitian. This proves uniqueness and local existence.
Under a holomorphic change of frame $e'=eg$, the metric matrix becomes $H'=g^\dagger Hg$. The <local formula for the Chern connection on a vector bundle> then gives
$$
A'=g^{-1}Ag+g^{-1}\partial g.
$$
This is precisely the transformation rule for a <connection on a vector bundle>, so the local connections glue and establish global existence. No Kähler hypothesis is needed for this part.
Extend $D$ to <vector-bundle-valued differential forms> by the graded Leibniz rule. The <curvature form of a connection> is the tensorial square $F_h=D^2$, acting by exterior multiplication. In the chosen frame,
$$
F_h=dA+A\wedge A=\bar\partial(H^{-1}\partial H).
$$
Indeed $\partial A+A\wedge A=0$ by differentiating $H^{-1}H=I$. It follows that $F_h$ has type $(1,1)$; the gauge change is $F'_h=g^{-1}F_hg$. Thus it is a global smooth two-form with values in $\operatorname{End}(E)$, namely an element of $\mathcal A^2(\operatorname{End}(E))$.
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