Prove the existence of an analytic function

Consider the function $f_1(z) = \frac{f(z^2)}{z^2}$. It is holomorphic on $\Delta$ and has no zeroes, so it will have a square root $h(z)$. Since $f_1$ is even, $h$ will be odd or even. Indeed, $h^2$ even implies $(h(z)-h(-z))(h(z)+ h(-z) )\equiv 0$. Since $h$ has no zeroes, it will be even. Define then $g(z) = z h(z)$. $g$ will be an odd function. You can check now that $g(z)^2 = f(z^2)$. Assume now $g(z_1) = g(z_2)$. Then $f(z_1^2) = f(z_2^2)$, and so $z_1^2 = z_2^2$. If $z_1 \ne z_2$, this implies $z_1 = - z_2$, and so $g(z_1) = - g(z_2)$. So both of them are $0$, $f(z_1^2) = 0$, so $z_1= 0 = z_2$. In conclusion, $g$ is injective and odd ( there are two solutions for $g$, clearly).

You want $g$ to be a branch of $\sqrt{f(z^2)}$. You have shown this to be possible in a neighbourhood of $0$. It is certainly possible in a neighbourhood of any other point of $\Delta$ (since $f$ is one-to-one and $f(0)=0$, $f(z) \ne 0$ there). The Monodromy Theorem should show that you get analytic continuation to $\Delta$. Moreover there are exactly two possible $g$ (since there are two choices for the square root in a neighbourhood of some $z \ne 0$).

Now if $g(z)$ is one solution, so are $-g(-z)$ and $g(-z)$. Thus one of these must be $g(z)$, i.e. $g$ is either even or odd. But by looking at the order of the zero at the origin, it can't be even.