If $f(z)$ is continuous on some domain $D$ and $f(z)^8$ (the function to the eighth power, not the eighth derivative) is analytic, then why does this imply that f is analytic on a neighborhood of each $z_0$ where $f(z_0)\not= 0$? Does this have to do with every branch cut of $z^a$ must go through the origin?
My reasoning is that at each point $z_0,f(z_0)\not=0,\in C$ an appropriate branch cut can be made so $z^{1/8}$ is defined there (just square root four times) and so analytic at each individual point where $f(z_0)\not=0$. Since $f$ is continuous, would this imply there is an entire neighborhood where this is true somehow? And then these zeros would be isolated, right?
Claim: If $f:\Omega\rightarrow\mathbb{C}$ is continuous. and $f^2\in H(\Omega)$. Let $C:=\{z\in \Omega| f(z)=0\}$ ($C$ is closed because $f$ is continuous), then $f|\Omega-C\in H(\Omega-C)$.
Proof: Let $z_0\in \Omega-C$. $\lim_{z\rightarrow z_0}\frac{f(z)^2-f(z_0)^2}{z-z_0}$ exists (Call this limit $L$). Using the continuity of $f$ and the fact that $f(z_0)$ is non-zero, one can show that $\lim_{z\rightarrow z_0}\frac{1}{f(z)+f(z_0)}$ exists as well. Hence, $\lim_{z\rightarrow z_0}[\frac{f(z)^2-f(z_0)^2}{z-z_0}\frac{1}{f(z)+f(z_0)}]$ and we get: $$\lim_{z\rightarrow z_0}[\frac{f(z)-f(z_0)}{z-z_0}]=\lim_{z\rightarrow z_0}[\frac{f(z)^2-f(z_0)^2}{z-z_0}\frac{1}{f(z)+f(z_0)}]=\,\,\,$$$$lim_{z\rightarrow z_0}\frac{f(z)^2-f(z_0)^2}{z-z_0}\lim_{z\rightarrow z_0}\frac{1}{f(z)+f(z_0)}=\frac{L}{2f(z_0)}$$
Hence $f|\Omega-C$ is analytic on $\Omega-C$ $\square$
Now going back to your question: Since $f$ is continuous, therefore $f^4$ is continuous. Since $(f^4)^2=f^8$ is analytic. By the previous claim, we get $f^4$ is analytic.
Since $f$ is continuous, therefore $f^2$ is continuous. Since $(f^2)^2=f^4$ is analytic. By the previous claim, we get $f^2$ is analytic.
We use the previous claim, for the third time to deduce that $f$ is analytic (because $f$ is continuous and because $f^2$ is analytic as we showed earlier) $\square$