Let $E \to M$ a Hermitian holomorphic vector bundle of rank $m$ over an almost-$\mathbb{C}$ hermitian Kähler manifold $(M,h)$ of dimesnion $n$. By a well known theorem $E \to M$ has a unique canonical hermitian holomorphic connection $\nabla^E$.
Identifying $TM$ with $T^{1,0}M$ from the Hodge decomposition of the complexified tangent space $TM \otimes \mathbb{C}= T^{1,0}M \oplus T^{0,1}M$. we choose (since the problem will be of local nature for appropriate open chart $U \subset M$) a orthonormal basis $(w_j)^n_{j=1}$ of $T^{1,0}M$.
Denote by $(\overline{w_j})^n_{j=1}$ the antiholomorphic corresonding basis of $T^{0,1}M$.
We will deal with $3$ operators: Dolbeault, Hodge and $\iota_X(-)$. Let us recall how each of them acts on $\mathcal{A}^{p,q}(M)$ resp. $\mathcal{A}^{p,q}(M,E)$:
Recall firstly, the Dolbeault operator $\overline{\partial}: \mathcal{A}^{p,q}(M) \to \mathcal{A}^{p,q+1}(M) $ is locally for a $\alpha:= a(x) \cdot w_I \wedge \overline{w_J} \in \mathcal{A}^{p,q}(U) $ with $I, I \subset \{1,...,n\}$ such that $ a(x) \in C^{\infty}(M,\mathbb{R})$ and $\vert I \vert =p$ und $\vert J \vert =q$ describing the holomorphic resp. antiholomorphic components of $\alpha$
defined by $\alpha \mapsto = \overline{\partial} \alpha := \sum_{j=1} ^n \frac{\partial a(x)}{\partial \overline{w_j}} \cdot \overline{dw_j} \wedge w_I \wedge \overline{w_J}$.
$\overline{\partial}$ extends naturally to $\mathcal{A}^{p,q}(M,E)$ by setting again locally for appropriate open chart $U \subset M$ an element $t:= \sum^m _{i=1} \alpha_i \otimes e_i \in \mathcal{A}^{p,q}(M,E) \vert _U = \mathcal{A}^{p,q}(U) \otimes \mathbb{R}^m$ as $\overline{\partial} t:=\sum_i \overline{\partial} \alpha_i \otimes e_i$
The Hodge operator $\ast: \mathcal{A}^{p,q}(M) \to \mathcal{A}^{n-p,n-q}(M)$ acts on $\alpha:= a(x) \cdot w_I \wedge \overline{w_J}$ by
$$\ast \alpha:= a(x) \cdot w_{I^C} \wedge \overline{w_{J^C}}$$
where $I^C=\{1,...,n\} \backslash I$ is the complement.
This again naturalally extends to $\mathcal{A}^{p,q}(M,E)$ in canonical way acting only on the $\mathcal{A}^{p,q}(M)$-component of the tensor.
The $\overline{\ast}$ is defined by $\overline{\ast} \alpha := \overline{\ast \alpha}$ (ie composition of $\ast$ with complex conjugation.
At last, the rejuvenation opertator (or contraction op) $\iota_X$ is defined in dependence of a vector field $X \in C^{\infty}(M, TM \otimes \mathbb{C})$ as follows:
Let again $\alpha:= a(x) \cdot w_{I^C} \wedge \overline{w_{J^C}}$ then $\iota_X \alpha :=\operatorname{Tr}_{11}( X \otimes \alpha)= a(x) \cdot w_{i_1}(X) \cdot w_{i_2} \wedge ... \wedge w_{i_p} \wedge \overline{w_J}$.
Recall, the notation $w_I$ abbreviates $w_{i_1} \wedge w_{i_2} \wedge ... \wedge w_{i_p}$ .
Now we come to my QUESTION:
Why the identity
$$\sum_{j=1} ^n \iota_{\overline{w_j}} \circ \nabla^E _{w_j}= \overline{\ast} \circ \overline{\partial} \circ \overline{\ast} $$
is true?
The local calculation for the right hand side for a $t:=\sum_i \overline{\partial} \alpha_i \otimes e_i \in \mathcal{A}^{p,q}(M,E) \vert _U$ is linear, so assume wlog $t= \alpha \otimes e_i$ with $\alpha:= a(x) \cdot w_{I} \wedge \overline{w_{J}}$.
We obtain
$$\overline{\ast} \circ \overline{\partial} \circ \overline{\ast}(a(x) \cdot w_{I} \wedge \overline{w_{J}} \otimes e_i)= \overline{\ast} \circ \overline{\partial} (a(x) \cdot \overline{w_{I^C}} \wedge w_{J^C} \otimes e_i) \\= \overline{\ast}(\sum_{j=1} ^n \frac{\partial a(x)}{\partial \overline{w_j}} \cdot \overline{dw_j} \wedge \overline{w_{I^C}} \wedge w_{J^C} \otimes e_i) \\= \sum_{j=1} ^n \frac{\partial a(x)}{\partial \overline{w_j}} \cdot w_{I \backslash \{j\}} \wedge \overline{w_{J}} \otimes e_i)$$
The left part is more complicated and I don't know ho to dealt with it, especially the $\nabla^E$-part. The problem is by definition locally a connection looks like $d+ \theta$ with $d$ the common differential and $\mathcal{A}^1(U, \mathbb{R}^{m \times m})$ and in contrast to the right hand side therefore "something happens" with $e_i$ compoents of the tensor product. I can't see what concretly. At least non enough to compare the calculation for the left hand side with above calculated right hand side.