What can be said about a complex polynomial $f$ with $\frac{\partial}{\partial \bar z} f^2 = 0$? The hypothesis is similar to that of a previous question of mine, namely that we have a complex polynomial $f: \mathbb{C} \to \mathbb{C}$ with  $\dfrac{\partial}{\partial \bar z} f^2 = 0$ . (In the previous question, $f$ was not squared, and the partial w.r.t. $z$ was also $0$)

What can be said about f?

At first glance, it seems that the square of $f$ doesn't depend on $\bar z$ (i.e. can be written in terms of $z$ alone). Is that all there is to this question? Are there some analytic/computational equivalent statements that I should be considering instead?

Added: I have tried to explicitly calculate $f^2$ and take its partial with respect to $\bar z$, but that doesn't seem to get me anywhere. Is there a less computational approach?
 A: Nothing can be said.
All complex polynomials, and their squares, are complex differentiable and therefore have zero $\frac{\partial}{\partial\overline z}$.
A: Well, if a complex-valued function $g$ satisfies $\dfrac{\partial}{\partial \bar z} g = 0$, we know that $\dfrac{\partial g}{\partial x} + i\dfrac{\partial g}{\partial y} = 0$ (nearly exactly from the definition of $\dfrac{\partial}{\partial \bar z}$) and so with a small bit of manipulation, we see that $i(\dfrac{\partial g}{\partial x} + i\dfrac{\partial g}{\partial y})  = i\dfrac{\partial g}{\partial x} - \dfrac{\partial g}{\partial y} = 0$ and $\dfrac{\partial g}{\partial y} = i\dfrac{\partial g}{\partial x}$. 
This is an equivalent formulation of the Cauchy-Riemann equations, which tells us our function $g$ is holomorphic and equivalently analytic.  In fact, a convenient test for holomorphicity is to check if $\dfrac{\partial}{\partial \bar z} = 0$.  So we know that our function $f^2$ is holomorphic, but all polynomials (and their squares, of course) are naturally holomorphic! Hence we could have concluded $\dfrac{\partial}{\partial \bar z}f = 0$ from our original hypothesis.  We have not actually constrained our function any further.
EDIT: Given your previous approach, a helpful fact to establish is that the product rule is valid for $\dfrac{\partial}{\partial \bar z}$.  Then $\dfrac{\partial f^2}{\partial \bar z} = 2f\dfrac{\partial f}{\partial \bar z}$, and so if $\dfrac{\partial f^2}{\partial \bar z} = 0$ and our polynomial isn't $0$ everywhere, then $\dfrac{\partial f}{\partial \bar z} = 0$ and we can write $f$ solely in terms of $z$, hence we can do the same for $f^2$. 
