If $f(z)^3$ is analytic then $f(z)$ is analytic.Is it true?if yes prove it.otherwise give counterexample. If $f(z)^3$ is analytic on $\Bbb{C}$ then $f(z)$ is analytic.Is it true?if yes prove it.otherwise give counterexample. 
 A: Since there is no continuity requirement for $f(z)$, just pick your favourite non-zero analitic function $g(z)$ and solve $f(z)^3=g(z)$. At almost all points you have three choices and you can easily make it non analytic.
E.g. Let $1, \omega, \omega^2$ be the three roots of unity. Define $f$ to be sometimes $1$, sometimes $\omega$... Then $f$ cannot be analytic, but $f^3$ is constant....
A: N.S. answer is from far the best possible answer here. But I wrote this and it works, so I guess I'll leave it here.
Let $\log$ be the principal branch of the complex logarithm. It is analytic on $\mathbb{C}\setminus (-\infty,0]$ and does not admit any analytic continuation beyond that.
Now set
$$
f(z):=e^{\frac{\log z}{3}}.
$$
Again, this is analytic on $\mathbb{C}\setminus (-\infty,0]$ and does not admit any analytic continuation beyond that. Not even a continuous extension (except at $0$).
For instance, you have
$$
\lim_{\theta\rightarrow \pi^+}\log(e^{i\theta})=i\pi\qquad \lim_{\theta\rightarrow \pi^-}\log(e^{i\theta})=-i\pi.
$$
So
$$ 
\lim_{\theta\rightarrow \pi^+}f(e^{i\theta})=e^{i\pi/3}\neq e^{-i\pi/3}=\lim_{\theta\rightarrow \pi^-}f(e^{i\theta}).
$$
So there is no analytic continuation at $-1=e^{i\pi}$.
Yet 
$$
f(z)^3=z
$$
admits an analytic continuation on $\mathbb{C}$.
