If $f : D(0,1) \rightarrow \mathbb{C}$ is a function, $f^2$ is holomorphic, and $f^3$ is holomorphic, then prove that $f$ is holomorphic.


If $f^3$ is holomorphic, then we can say that $\frac{\partial}{\partial z}(f^3)$ must be holomorphic. Since the chain-rule for differentiation still applies with complex-valued functions, we can say that

$$\frac{\partial}{\partial z}(f^3) = 3f^2\left(\frac{\partial f}{\partial z}\right) \implies 3f^2\left(\frac{\partial f}{\partial z}\right) \textrm{ is holomorphic}$$

Now it may be unclear whether this implies that $\frac{\partial f}{\partial z}$ is holomorphic even knowing that $f^2$ is holomorphic. Certainly the product of two holomorphic functions must be holomorphic, but the case in which one is holomorphic and the other isn't but the product of those two function is holomorphic isn't as clear, so it will be shown.

Suppose we have two functions, $g$ and $h$. Suppose $g$ is holomorphic on some open subset $\Omega \subset \mathbb{C}$. Now suppose their product, defined as $\eta = g\cdot h$, is holomorphic. This means that $\eta \cdot (1/g)$ is holomorphic except at those points in $\Omega$ where $g$ is zero. Thus, $h$ must be also holomorphic at all points except those in which $g$ is zero, so $h$ is at least meromorphic.

Using that, we can see that $\frac{\partial f}{\partial z}$ must be at least meromorphic, being holomorphic at all points with no guarantee as to whether it has singularities at those points where $f^2$ is zero.

And that's where I'm stuck. I know that $\frac{\partial}{\partial z}(f^3)$ and $3f^2$ are related, but as long as the possibility that $\frac{\partial f}{\partial z}$ blows up at those points where $f^2\rightarrow 0 $ has to be considered, I'll stay confused.

Is my approach alright, or should I completely change what I'm doing? Have I gone wrong anywhere? Should I be more explicit anywhere? Could you give me a small minor hint as to which direction I should go?