If $f(z)$ is defined in the unit disk such that both $f^2(z)$ and $f^3(z)$ are analytic in the disk. is $f(z)$ analytic in the disk?
Attempt: If $f(z) \ne 0$ in the disk, then $f(z) = \dfrac{f^3(z)}{f^2(z)}$ must be analytic.
Hence, any counterexample involving $f$ must be such that $f(z) = 0$ at least one point in the disk.
Any hints on how to to move forward from here?
If $f$ is the zero function then the statement is obviously true. Otherwise, $f^2$ is a nonzero holomorphic function, and so its zeros are isolated. Equivalently, the zeros of $f$ are isolated. The function $f$ is clearly analytic at all the other points of the disk, and its zeros are isolated singularities. So in order to show that $f$ is analytic at the whole disk, you just have to show that these singularities are removable. And this is indeed the case. If $f$ had a pole at such a point then so would $f^2$. If $f$ had an essential singularity then using Casorati-Weierstrass (or Picard's theorem if you know it) it is easy to show that $f^2$ would also have an essential singularity there. Both contradict the assumption that $f^2$ is holomorphic.