= Solution
On a <complex manifold>, the <exterior derivative> splits by type as $d=\partial+\bar\partial$. The <conjugate Dolbeault operator> $\partial$ raises <holomorphic> degree by one and the <Dolbeault operator> $\bar\partial$ raises antiholomorphic degree by one. Their identities are
$$
\partial^2=\bar\partial^2=0,\qquad\partial\bar\partial+\bar\partial\partial=0.
$$
Use the convention $d^c=i(\bar\partial-\partial)$ for the <d c operator>; then $dd^c=2i\partial\bar\partial$. The alternative convention with a factor $1/2$ changes this last factor only. <Complex conjugation of differential-form type> interchanges $\partial$ and $\bar\partial$ without reversing the wedge order. Thus, for a real form $\eta$,
$$
\overline{i\partial\bar\partial\eta}=-i\bar\partial\partial\eta=i\partial\bar\partial\eta.
$$
\b[The resulting form is real], independently of the degree or type components of $\eta$.
The <Dolbeault cohomology> is
$$
H^{p,q}(X)=\frac{\ker(\bar\partial:\mathcal A^{p,q}\to\mathcal A^{p,q+1})}{\operatorname{im}(\bar\partial:\mathcal A^{p,q-1}\to\mathcal A^{p,q})}.
$$
In degree $(0,0)$ there is no incoming image, and the kernel consists of global <holomorphic functions>. On a <compact> <connected> $X$, $|f|$ attains a maximum. In a coordinate polydisc around a maximizing point, the one-variable <maximum modulus principle> applied successively along the coordinate directions makes $f$ constant on a smaller polydisc. Alternatively the same argument shows the maximum set is open; it is closed and nonempty, so <connectedness> makes it all of $X$, and the local constancy then makes $f$ globally constant. Consequently
$$
\boxed{H^{0,0}(X)\cong\mathbb C.}
$$
A <holomorphic p-form on a complex manifold> is a form of type $(p,0)$ with <holomorphic coordinate> coefficients, equivalently one killed by $\bar\partial$. For degree $n=\dim_{\mathbb C}X$, it is $d$-closed: its $\bar\partial$ vanishes by holomorphicity and its $\partial$ has impossible bidegree $(n+1,0)$. Its conjugate is also closed.
The key positivity behind <nonzero holomorphic top forms are not exact> is explicit. If $\alpha=f\,dz_1\wedge\cdots\wedge dz_n$, complex orientation gives
$$
i^{n^2}\alpha\wedge\bar\alpha
=2^n|f|^2\,dx_1\wedge dy_1\wedge\cdots\wedge dx_n\wedge dy_n.
$$
This is nonnegative everywhere and positive on some open set when $\alpha\not\equiv0$, so its integral on the <compact> manifold is strictly positive. If $\alpha=d\gamma$, closedness of $\bar\alpha$ and the <Stokes theorem> would give
$$
\int_X\alpha\wedge\bar\alpha=\int_Xd(\gamma\wedge\bar\alpha)=0.
$$
Here Stokes is used on a <compact> oriented manifold without boundary, and complex-valued forms are handled by linearity. The contradiction proves \b[a nonzero <holomorphic> top form is not $d$-exact]. For dimension zero the same nonexactness follows directly from the absence of degree $-1$ forms.
Now take a <compact> <complex surface>, with no <Kähler metric> assumed. For a <holomorphic> one-form $\alpha$, the form $\beta=d\alpha=\partial\alpha$ has type $(2,0)$ and is <holomorphic>, since $\bar\partial\partial\alpha=-\partial\bar\partial\alpha=0$. It is exact. The top-form result forces $\beta=0$, proving closedness. This establishes <holomorphic forms on a compact complex surface are closed> by an argument which does not require a Kähler identity.
If a <holomorphic> one-form were exact, write $\alpha=df$ for a global smooth complex function. Comparing types gives $\bar\partial f=0$. The preceding <compactness> argument makes $f$ constant on every <connected> component, hence $\alpha=0$. <Holomorphic> two-forms are closed and nonexact unless zero by the top-form result. <Holomorphic> zero-forms are constant on each component, and a nonzero zero-form is never exact because there are no negative-degree forms. Degrees above two contain only zero forms. Therefore \b[every <holomorphic> form on a <compact> complex surface is closed, and a nonzero one is never exact].
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