$X$ is a compact Riemann surface. $\frac{\wedge^2(X)}{(\partial\bar{\partial}C^{\infty}(X)}\cong C$. This is an exercise 19 b) of Forster Lectures on Riemann Surface Chpt 2, Sec 19.
$H^1(X,\mathcal{H})\cong\frac{\wedge^2(X)}{(\partial\bar{\partial}C^{\infty}(X)}\cong C$ where $\mathcal{H}$ is sheaf of harmonic functions over $X$. It is clear by $0\to\mathcal{H}\to C^{\infty}(X)\xrightarrow{\partial\bar{\partial}}\wedge^2(X)\to 0$ is short exact sequence.(One needs to invoke Dolbeault lemma and use simply connectedness of disk for surjectivity on stalk level.)
Denote $\wedge^2(X)$ the smooth 2 forms over $X$. Basically I want to write $\wedge^2(X)=Harm(X)\cdot Vol(X)\oplus\partial\bar{\partial}C^{\infty}(X)$ where $Harm(X)$ is the harmonic functions, $Vol(X)$ is the volume form, $\partial=\frac{\partial}{\partial z}$ and similarly for $\bar{\partial}$.
Since $X$ is compact, every harmonic function is constant. I will have isomorphism. 
$\textbf{Q:}$ How should I show above isomorphism of $\frac{\wedge^2(X)}{(\partial\bar{\partial}C^{\infty}(X)}\cong C$? Since the book did not discuss extending $L^2$ inner product to 2-forms, I do not know how to progress. If I assume extending $L^2$ innter product to 2-forms, then it is clear that $Harm(X)\cdot Vol(X)\oplus\partial\bar{\partial}C^\infty(X)\subset\wedge^2(X)$. Naively speaking, I need codifferntial send 2 forms to functions and remove constant $Harm(X)$ part. This could be potentially done by $d$. Then I do not see how to proceed.
$Update:$ I think I can figure out the constant part if I assume hodge decomposition as preassumption. Consider $\omega\in\wedge^2(X)$. Take normalized volume $vol$ of $X$ as it is orientable. Then $\int_X\omega=c$. Consider $\omega-c\cdot vol$. Now one needs to show $\omega-c\cdot vol\in\partial\bar{\partial}C^{\infty}(X)$.
 A: Let $X$ be a compact Riemann surface.
By theorem $19.9$ of Lectures on Riemann Surfaces by Otto Forster and Bruce Gilligan, we get
$$\mathcal{E}^{0,1}(X)=\overline{\partial}\mathcal{E}(X)\oplus\overline{\Omega}(X).$$
Then take conjugation, we get
$$\mathcal{E}^{1,0}(X)={\partial}\mathcal{E}(X)\oplus{\Omega}(X).$$
Then
$$\overline{\partial}\mathcal{E}^{1,0}(X)={\partial}\overline{\partial}\mathcal{E}(X),$$
since $\overline{\partial}\partial=-\partial\overline{\partial}$ and  $\overline{\partial}{\Omega}(X)=0$. Now, we can say
$$H^1(X,\mathcal{H})\cong \mathcal{E}^2/{\partial}\overline{\partial}\mathcal{E}(X)=\mathcal{E}^2/\overline{\partial}\mathcal{E}^{1,0}(X)=H^{1,1}(X).$$
Fact.  Let $X$ be a compact Riemann surface. Then we have a linear isomorphism
$$H^2(X,\mathbb{C})\cong H^{1,1}(X).$$
This fact follows from the following theorem called "$\partial\overline{\partial}$-lemma":
Theorem. ($\partial\overline{\partial}$-lemma) Let $X$ be a Riemann surface, and $\mu$ be a $2$-form on $X$ with compact support. Suppose that
$$\int_X\mu=0.$$
Then there exists $f$ in $\mathcal{E}(X)$, such that
$$\mu=d*df=2i\partial\overline{\partial}f.$$
Then $H^1(X,\mathcal{H})\cong H^{2}(X,\mathbb{C})\cong \mathbb{C}$, since $X$ is compact.
