= Solution
Use both <Kähler identities> from the preceding parts, so that $\partial^*=i[\Lambda,\bar\partial]$ and $\bar\partial^*=-i[\Lambda,\partial]$. For odd operators $a,b$, write $\{a,b\}=ab+ba$. Direct expansion gives
$$
\{\partial,[\Lambda,\bar\partial]\}+\{\bar\partial,[\Lambda,\partial]\}
=[\Lambda,\{\partial,\bar\partial\}]=0.
$$
Since $\partial\bar\partial+\bar\partial\partial=0$, we obtain
$$
\Delta_\partial=i\{\partial,[\Lambda,\bar\partial]\}
=-i\{\bar\partial,[\Lambda,\partial]\}=\Delta_{\bar\partial}.
$$
Together with part (b), this proves the full <Kähler Laplacian identity>
$$
\boxed{\Delta_d=2\Delta_\partial=2\Delta_{\bar\partial}.}
$$
Now take the exact form $\eta$. Being $\bar\partial$-exact, it is orthogonal to the harmonic space, so $H\eta=0$, and it is also $\bar\partial$-closed. Choose $\alpha=G\eta$ for the <Dolbeault Green operator> of the <Dolbeault Laplacian>. Commutation gives $\bar\partial\alpha=G\bar\partial\eta=0$. Therefore
$$
\boxed{\eta=\Delta_{\bar\partial}\alpha=\bar\partial\bar\partial^*\alpha.}
$$
Suppose also $\partial\eta=0$, and put $v=\bar\partial^*\alpha$, $u=\partial v$. The mixed identities and $(\bar\partial^*)^2=0$ give
$$
\bar\partial u=-\partial\bar\partial v=-\partial\eta=0,\qquad
\bar\partial^*u=-\partial\bar\partial^*v=0.
$$
Thus $u=\partial\bar\partial^*\alpha$ is harmonic for the <Dolbeault Laplacian>. By the <Kähler Laplacian identity> it is also $\partial$-harmonic and hence $\partial^*u=0$. But it is $\partial$-exact, so
$$
\|u\|^2=\langle\partial v,u\rangle=\langle v,\partial^*u\rangle=0.
$$
Consequently \b[$\boxed{\partial\bar\partial^*\alpha=0}$]: the form $v$ is $\partial$-closed.
For every harmonic form $h$ of the same bidegree as $v$, $\langle v,h\rangle=\langle\alpha,\bar\partial h\rangle=0$. Harmonic spaces for the two Dolbeault Laplacians coincide, so $v$ has no $\partial$-harmonic component. Apply the $\partial$ version of the <Dolbeault Hodge decomposition>. Since $v$ is $\partial$-closed, it has no $\partial^*$-coexact component either. Thus $v=\partial\phi$, with $\phi\in A^{p-1,q-1}$. We conclude the requested <ddbar lemma>:
$$
\boxed{\eta=\bar\partial\partial\phi.}
$$
For example $\phi=\partial^*Gv$ is a choice, since the same Green operator serves $\Delta_\partial=\Delta_{\bar\partial}$. If $p=0$ or $q=0$, the relevant negative-bidegree space is zero and the argument forces $\eta=0$. Compactness, the absence of boundary and the <Kähler> condition are essential to this global harmonic argument.
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