TITLE: Zero of the derivative of ameromorphic function
QUESTION [5 upvotes]: Let $f:\mathbb{C} \rightarrow \hat{\mathbb{C}}$ a meromorphic function with an essential singularity at infinity. Does $f'(z)$ have a zero? No, considering $z\mapsto e^z$. But if I assume that $f$ is surjective, is it true?  and can i say something on the order of vanishing.
I am not a specialist of holomorphic function. This question arise in geomerty. More precisely my function satisfies $\frac{f'}{1+\vert f\vert^2}=O(\frac{1}{z})$, where the right-hand side is called the spherical derivative. 
Thanks to the work of Letho,The spherical derivative of meromorphic functions in the neighborhood of an isolated singularity, we know that $f$ must be surjective.
Added: The full question is: If $f$ has an essential singularity and  satisfies $\frac{f'}{1+\vert f\vert^2}=O(\frac{1}{z})$, does $f'$ vanishes? and can we say something about the order of vanishing.

REPLY [3 votes]: A simple counter-example is:
$$f(z)=2\int_0^ze^{w^2}dw-\frac{e^{z^2}}{z}.$$
Obviously $f$ is a meromorphic function on $\Bbb C$ with an essential singularity at infinity and 
$$f'(z)=\frac{e^{z^2}}{z^2}$$ 
has no zero. 
To show $f(\Bbb C)=\hat{\Bbb C}$, since $f(0)=\infty$, let us assume that there exists $c\in \Bbb C$ such that $c\notin f(\Bbb C)$ to get some contradiction. By our assumption, 
$$g(z):=z(f(z)-c)=2z\int_0^ze^{w^2}dw-e^{z^2}-cz$$
is an entire function with no zero. Moreover, it is easy to see that the order of $g$ at infinity is $2$. Then by Hadamard's theorem, there exists a polynomial $P$ of degree $2$, such that
$$g(z)=e^{P(z)}\Longrightarrow g'(z)=P'(z)g(z),$$
which is absurd.