Disproving the statement "If $f(z)$ is not an entire function, then $g(z) = f^2(z)$ is not entire.” 
If $f(z)$ is not an entire function, then $g(z) = (f(z))^2$ cannot be an entire function.”

From this statement why would saying "Let $f(z) = \sqrt z$, assuming that it is a branch that takes $−1$ to $i$. Then it is not analytic on the branch cut, but $g(z) = (f(z))^2 = z$ is obviously an entire function."
Not be a proper example to disprove this statement/ be a incorrect counterexample?
I understand that for $\sqrt z$ to be a incorrect counterexample means it is not analytic on the branch -1 to I, therefore not entire.
However I thought $\sqrt z$ would be analytic throughout the whole branch cut from -1 to i. Since on the complex plane,
$\lim_{z \to -1} \sqrt z$ exists as it would approach $i$ in that case, and in the other case $\lim_{z \to i} \sqrt z$
it approaches $\sqrt i$ meaning that at least on that branch cut in the complex plane it is continuous everywhere, therefore the partial derivatives exist everywhere on the branch cut meaning it is analytic, but this assumption seems to be incorrect. What am I missing or not understanding here.
Precisely, I don't understand why $\sqrt z$ would not be a proper counterexample to the statement.
 A: Your counterexample is essentially correct, although when one talks about a "branch" of the square root function, one generally means an open set $U \subseteq \mathbb{C}$ together with a holomorphic function $f : U \to \mathbb{C}$ such that $(f(z))^2 = z$ for all $z \in U$. In this case, that is not what we want.
In fact, all you need to know is that for all $z \in \mathbb{C}$, there exists a $p \in \mathbb{C}$ such that $p^2 = z$. Then there must be some "square root function" $f$ which satisfies $(f(z))^2 = z$ for all $z$. But this function cannot be entire because it cannot be differentiable at $0$, even though its square is entire.
A: Your example IS a counterexample to the statement above. Yes, it is possible to have a function that cannot be made entire by analytic extension such that the square is an entire function. The square root is precisely such an example.
We can make this more concrete. We will define "a" square root function to be based on the polar representation of each point in $\mathbb{C}$; i.e., it is well known that given a complex number $z\not=0$ there exist unique $r\in\mathbb{R}^+$ and $\theta\in[-\pi,\pi)$, such that $z=re^{i\theta}$. Hence, we define $f(0)=0$ and for $z=re^{i\theta}\not=0$,
$$f(z)=\sqrt{r}e^{i\theta/2}$$
We see that $f(z)$ is defined and exists for every value in $\mathbb{C}$. Moreover, $f(z)^2=z$ for every value in $\mathbb{C}$. However, $f(z)$ (while defined at every negative real value) is not continuous at any value on the negative real axis. This is because a limit from the 2nd quadrant down towards a point and a limit from the 3rd quadrant up towards a point, will not agree. Explicitly, if $a\in(-\infty,0)$, then
$$\lim_{\epsilon\to 0}f(a+\epsilon i)=\lim_{\epsilon\to 0}\sqrt{-a-\delta_\epsilon}e^{i(\pi-\gamma_\epsilon)/2} = \sqrt{-a}e^{i(\pi/2)} = i\sqrt{-a} $$
but
$$\lim_{\epsilon\to 0}f(a-\epsilon i)=\lim_{\epsilon\to 0}\sqrt{-a-\delta_\epsilon}e^{i(-\pi+\gamma_\epsilon)/2} = \sqrt{-a}e^{i(-\pi/2)} = -i\sqrt{-a}$$
These two values are not the same value and hence no general limit can exist at the value of $a$. This means $f$ fails to be continuous at $a$ and means that $f$ fails to be analytic at $a$. Everywhere, (besides $z=0$ and the explicitly chosen interval endpoints of $[-\pi,\pi)$) the function is analytic. Moreover we could have chosen any interval we wanted, thus creating a new "branch cut"; hence there are many possible square root functions that can be defined.
