= Solution
The <Kähler manifold> structure gives a <Riemannian metric>, its volume form, the complex orientation, and a Hermitian inner product on complex differential forms. The <complex Hodge star operator> is the complex-linear map characterized by
$$
\alpha\wedge *\bar\beta
=\langle\alpha,\beta\rangle\operatorname{vol}_g.
$$
On $r$-forms in real dimension $m=2n$, the <codifferential> is
$$
d^*=(-1)^{m(r+1)+1}*d*,
$$
equivalently the <formal adjoint> of $d$ for the $L^2$ inner product. Similarly $\bar\partial^*$ is the formal adjoint of $\bar\partial$. Define the <Hodge Laplacian> and <Dolbeault Laplacian> by
$$
\Delta_d=dd^*+d^*d,
\qquad
\Delta_{\bar\partial}=\bar\partial\bar\partial^*+\bar\partial^*\bar\partial.
$$
Expanding $d=\partial+\bar\partial$, the <Kähler identities> make the mixed anticommutators vanish and imply $\Delta_\partial=\Delta_{\bar\partial}$. Hence the <Kähler Laplacian identity> is
$$
\Delta_d=2\Delta_{\bar\partial}.
$$
Let $L\gamma=\omega\wedge\gamma$ be the <Lefschetz operator of a Kähler manifold>. The Kähler identities also imply
$$
[L,\Delta_{\bar\partial}]=0.
$$
Thus, if $\Delta_{\bar\partial}\alpha=0$, then
$$
\Delta_{\bar\partial}(\alpha\wedge\omega^k)
=\Delta_{\bar\partial}L^k\alpha
=L^k\Delta_{\bar\partial}\alpha=0.
$$
This is the fact that the <Lefschetz operator preserves harmonic forms>.
The <Dolbeault Hodge decomposition on a compact Hermitian manifold> states that
$$
\Omega^{p,q}(X)
=\mathcal H_{\bar\partial}^{p,q}
\oplus\bar\partial\Omega^{p,q-1}(X)
\oplus\bar\partial^*\Omega^{p,q+1}(X),
$$
an orthogonal direct sum, where $\mathcal H_{\bar\partial}^{p,q}=\ker\Delta_{\bar\partial}$.
Suppose $\eta=\bar\partial\gamma$ has type $(p,q)$. Apply this decomposition to $\gamma\in\Omega^{p,q-1}$. The harmonic and $\bar\partial$-exact pieces disappear after applying $\bar\partial$, so for some $\beta\in\Omega^{p,q}$,
$$
\eta=\bar\partial\bar\partial^*\beta.
$$
Put $\theta=\bar\partial^*\beta$. If also $\partial\eta=0$, then
$$
\bar\partial(\partial\theta)=-\partial(\bar\partial\theta)=-\partial\eta=0.
$$
The Kähler anticommutation identity $\bar\partial^*\partial+\partial\bar\partial^*=0$ and $(\bar\partial^*)^2=0$ give
$$
\bar\partial^*(\partial\theta)
=-\partial(\bar\partial^*\theta)=0.
$$
Therefore $\partial\theta$ is $\Delta_{\bar\partial}$-harmonic. By $\Delta_\partial=\Delta_{\bar\partial}$ it is also $\partial$-harmonic, but it is $\partial$-exact; orthogonality of harmonic and exact forms forces
$$
\partial\theta=0.
$$
This proves both requested claims: $\partial\bar\partial^*\beta=0$ is harmonic, and $\bar\partial^*\beta=\theta$ is $\partial$-closed.
Finally, $\theta\in\operatorname{im}\bar\partial^*$ is orthogonal to $\ker\bar\partial$, and hence to every $\bar\partial$-harmonic form. Since the $\partial$- and $\bar\partial$-harmonic spaces agree on a compact Kähler manifold, the $\partial$-closed form $\theta$ has zero harmonic component in its $\partial$-Hodge decomposition. It follows that $\theta=\partial\phi$ for some $\phi\in\Omega^{p-1,q-1}(X)$. Hence
$$
\eta=\bar\partial\theta=\bar\partial\partial\phi,
$$
which is the <ddbar lemma> in this case.
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