$f^2$ and $f^3$ are holomorphic implies $f$ is holomorphic.

Suppose $f$ is a continuous complex valued function on a domain $\Omega$. Suppose $f^2$ and $f^3$ are holomorphic in $\Omega$. Show that $f$ is also holomorphic in $\Omega$.

Assume $f=u+iv$. I see that if $u,v$ are in $C^1$ then $f^2$ is holomorphic can imply $f$ is holomorphic considering Cauchy-Riemann equations. But I don't know how to get $u,v$ are in $C^1$ by the adding condition $f^3$ is holomorphic. (I can't get $u,v$ from some combinations of real and imaginary part of $f^2$ and $f^3$). Do someone know how to do this?

The zeros of $f^2$ are a closed discrete set by the identity theorem. On their complement, $f^3/f^2$ is defined, holomorphic and coincides with $f$.

Now use the continuity of $f$ to conclude that all the singularities of $f^3/f^2$ are in fact removable, proving $f = f^3/f^2$ on all $ℂ$.