YangMills' answer shows that it is not always possible to represent a real $(1,1)$-form $\phi$ in the desired form globally on a compact complex manifold but doesn't answer the question of how to tell, for a given $\phi$, whether it is possible locally.  Under mild conditions, though, it is possible to derive necessary and sufficient conditions.

For example, on the (possibly empty) open set where $\phi^2\not=0$, if there exists a $u$ such that $\phi = u\,\mathrm{i}\partial\bar\partial u$, then the identity 
$$
\phi^2 = u^2\,(\mathrm{i}\partial\bar\partial u)^2 = u^2(\mathrm{i}\partial\bar\partial\phi)\tag1
$$
implies that $\mathrm{i}\partial\bar\partial\phi\not=0$ and that the ratio $\phi^2 /( \mathrm{i}\partial\bar\partial\phi)$ must be a smooth positive function $f$.  Conversely, if $\mathrm{i}\partial\bar\partial\phi\not=0$ and there is a (necessarily unique) positive function $f$ such that $\phi^2 = f\,(\mathrm{i}\partial\bar\partial\phi)$, then $\phi$ is of the desired form if and only if $$\phi = \sqrt{f}\,\mathrm{i}\,\partial\bar\partial\left(\sqrt{f}\right).\tag2$$

On the (possibly empty) interior of the set on which $\phi^2=0$ but $\phi\not=0$,
there is a more elaborate method of testing whether $\phi$ can be written in the desired form, which starts by observing that the identity (1) now implies that $\mathrm{i}\,\partial\bar\partial\phi = 0$ and, moreover, because $\mathrm{i}\,\partial\bar\partial u$ is closed, it follows that $\phi$ must be a multiple of a closed *simple* $2$-form, and, hence, $\mathrm{d}\phi = \mu\wedge\phi$ for some real $1$-form $\mu$.  To go further requires more elaborate calculations, which I can supply if there is interest.  I will just note here that, in this case, the representation of $\phi$ in the desired form is often no longer unique up to sign.