I wrote a blog post about almost exactly this question. I'll give a summary here:

Since $H^{1,1}(X)$ is one dimensional, I could answer your question by giving anythng with the correct integral. However, I'll try to give you the kind of cocycle which actually comes out of the proof of Dolbeaut-Cech equality.

Your cocycle isn't $dz/z$ but, rather, $dz/z$ with a specific choice of open cover of $X$. Lets say your choice is $U_1 \cup U_2$, where $U_1 = \{ z : z \neq \infty \}$ and $U_2 = \{ z : z \neq 0 \}$. Refine your cover to $V_1 \cup V_2$, where $V_1 = \{ z : |z| > r \}$ and $V_2 = \{ z : |z| < r^{-1} \}$ for some $r < 1$.

Let $\theta_1$ and $\theta_2$ be $1$ forms on $V_1$ and $V_2$ such that $\theta_1|_{V_1} - \theta_2|_{V_2} = dz/z$. Then $\overline{\partial} \theta_1$ and $\overline{\partial} \theta_2$ have equal restrictions to $V_1 \cap V_2$. The $(1,1)$-form you are looking for is their common value, which I'll call $\omega$.

Let's first do a fake solution. **A real solution would look like a $C^{\infty}$ smearing out of this one.**

We'll work in the degenerate case $r=1$, so we are only gluing along a circle, not an annulus. We'll take $\theta_1 = (1/2) \ \overline{z}\ dz$ and $\theta_2 = -(1/2) \ dz / (\overline{z} z^2)$. Notice that both $\theta_1$ and $\theta_2$ restrict to $dz/z$ on the unit circle, but are constructed to extend smoothly to the appropriate discs.

So $\overline{\partial} \theta_1 = (1/2) d \overline{z} d z$ and $\overline{\partial} \theta_2 = (1/2) dz d \overline{z} / (\overline{z}^2 z^2)$. Our $\omega$ is formed by gluing these two differential forms together.

A **genuine smooth solution** would be like this, but would interpolate smoothly between these two. If you push forward in a brute force manner, you'll get something with bump functions in it.

If you are more clever, you may discover the solution
$$\theta_1 = \frac{dz}{z} \left( 1- \frac{1}{1+z \overline{z}} \right)$$
and
$$\theta_2 = - \frac{dz}{z} \left( \frac{1}{1+z \overline{z}} \right).$$

You should check that $\theta_1|_{V_1} - \theta_2|_{V_2} = dz/z$ and that $\theta_i$ is smooth and well-defined on $U_i$.

Then
$$\overline{\partial} \theta_1 = \overline{\partial} \theta_2 = \frac{dz \ d\overline{z}} {(1+z \overline{z})^2}.$$

This is, as Scott guessed, the **Fubini-Study** form.