holomorphic function with special decreasing property If you consider $f=\frac{P}{Q}$ the quotient of two polynomial function then $\frac{f'}{1+\vert f\vert^2}$ decrease like $\frac{1}{z}$. My question is, is the converse true? is an meromorphic function(define on the whole plane)  which satisfies
$$\frac{f'}{1+\vert f\vert^2}=O\left(\frac{1}{z}\right)$$
as $z$ goes to infinity, is the quotient of two polynomial function?
 A: EDIT. I have edited point 4) below to provide more details, and corrected the claim in Lehto's theorem.
This answer is positive if $f$ is entire, but negative for general meromorphic functions.
To simplify notation, let $f^{\#}(z)$ denote the spherical derivative, i.e. $f^{\#}(z) = |f'(z)|/(1+|f(z)|^2)$.
A little bit of further research on the web unsurprisingly showed that the result is not new. Indeed, this problem was introduced by Lehto and Virtanen. They prove most of what is needed for your question, but I don't have access to the paper. However, Lehto proved a stronger result in his paper "The spherical derivative of meromorphic functions in the neighbourhood of an isolated singularity", Comment. Math. Helv. 33, Number 1:
Theorem 1. If $f$ is meromorphic in a punctured neighborhood of infinity, and suppose that $f$ has an essential singularity at infinity. Then
$$\limsup |z| f^{\#}(z) \geq \frac{1}{2}.$$ This inequality is best possible: there are meromorphic functions on the punctured plane for which equality holds (and these can be described explicitly as certain infinite products). 
On the other hand, if $f$ is analytic (i.e. has no poles), then 
$$\limsup |z| f^{\#}(z) = \infty.$$
It may still be useful to give my account of (some weaker, but sufficient) versions of these results, particularly since the Lehto-Virtanen paper is not available online, and might or might not be in your local university library. Also, my account explicitly shows that you can get a function that is meromorphic in the whole plane which is a counterexample to your question, while, as you note, Lehto's example is not meromorphic at zero.
1) To prove the final statement in the theorem, we use the theory of normal families. Consider the functions
$$f_{\lambda}(z) := f(\lambda z); \quad |\lambda|\geq 1.$$
If $\limsup |z| f^{\#}(z) < \infty$, then it follows that the spherical derivative of $f_{\lambda}$ remains locally bounded near every point $z$ with $|z|>0$ as $\lambda\to\infty$. Hence these maps form a normal family by Marty's theorem. Suppose that $f$ is analytic, and let $z_n\to\infty$ be a convergent sequence; we may assume without loss of generality that the limit $z_0$ is finite (as otherwise $f$ has a pole at infinity and there is nothing to prove). By passing to a subsequence, we can assume that the maps $f_{z_n}$ converge locally uniformly to a limit function $g$, which is defined on the punctured complex plane. Since $f_{z_n}(1)=f(z_n)\to z_0$, we have $f(1)=z_0$; in particular, $g$ is not constantly equal to infinity, and hence $g$ is analytic. It follows that the maximum modulus of $f$ on the circle $|z|=|z_n|$ remains bounded as $n\to\infty$. By the maximum modulus principle, $f$ is bounded near infinity, and hence $f$ has a removable singularity at infinity. This completes the proof.
2) A similar argument should allow us to prove that 
$$\limsup |z| f^{\#}(z) >0$$
when $f$ is meromorphic and the singularity is not removable. (In this case, all limit functions have to be constant, which should lead to a contradiction. Certainly one can get a contradiction using the essential singularity version of the Ahlfors five islands theorem, but that's quite a sledgehammer. Lehto's proof, mentioned below, is much easier.) Of course once this is proved, we also see that there is a universal constant $c>0$ such that 
$$\limsup |z| f^{\#}(z) >c$$
(again using normal families).
3) Finally, to construct a function $f$ such that 
$$\limsup |z| f^{\#}(z) <\infty,$$
just take a meromorphic function $f:\mathbb{C}\setminus\{0\}\to\overline{\mathbb{C}}$ such that $f(\lambda z)=f(z)$ for some complex $\lambda$, $|\lambda|>1$, and all $z$. In other words, $f$ is of the form
$$f(e^z) = g(z),$$
where $g$ is an elliptic function with periods $2\pi i$ and $\log(\lambda)$, for example $g$ could be a Weierstrass $\wp$-function.
4) The above example is not meromorphic on the entire complex plane, but one could turn it into a function that is by using quasiconformal mappings. 
To do so, construct the function $f$ using a Weierstrass function on a rectangular lattice (so $\lambda\in (1,\infty)$). Let $C$ be a circle centered at the origin and not passing through critical points, and let $\gamma$ be the image of $C$ under $f$. The properties of the $\wp$-function imply that $f$ is conformal and one-to-one in a neighborhood of $C$, so $\gamma$ is an analytic Jordan curve. Let $U$ be the complementary domain of $\gamma$ on the Riemann sphere for which $\gamma$ is the boundary of $D$ described in positive orientation. (Choosing $C$ appropriately, we could make sure that $U$ is the bounded complementary component of $\gamma$, but this doesn't matter for our purposes.)
Let $\phi:U\to \mathbb{D}$ be a conformal isomorphism, where $\mathbb{D}$ is the unit disk. Such an isomorphism exists by the Riemann mapping theorem, and extends analytically to a neighborhood of $U$ because $\gamma$ is an analytic curve (Schwarz reflection). Let $r$ be the radius of the circle $C$; then the map
$$ \psi:S^1\to S^1; \quad z\mapsto \phi(f(rz))$$
is an orientation-preserving analytic circle diffeomorphism, and as such extends to a quasiconformal homeomorphism $\psi:\mathbb{D}\to\mathbb{D}$. (You can just define such an extension by hand, using $\psi$ as a function on the argument and fixing the modulus, or use something like the Douady-Earle extension.) 
Now define
$$\tilde{h}:\mathbb{C}\to \overline{\mathbb{C}}; \quad
z\mapsto \begin{cases} f(z) & |z|\geq r \\ 
\phi^{-1}\left(\psi\left(\frac{z}{r}\right)\right) & |z|<r.\end{cases}$$
This is a quasiregular map. Using the measurable Riemann mapping theorem, we can write 
$\tilde{h}= h\circ\theta$, where $\theta$ is meromorphic and $\phi$ is a quasiconformal homeomorphism that is conformal near infinity. It follows that $h$ has the desired property. 
This leaves open the question of whether Lehto's sharp examples can be realized (either exactly or asymptotically) by a function meromorphic in the complex plane. I suspect that this can be done by a similar method as above, but I have not thought about this.
5) Lehto's proof of the first statement in Theorem 1 above is actually very simple and elegant (though somewhat miraculous): Consider the function
$$F(z) := f(z)\cdot \overline{f}(\overline{z}e^{2\pi i \theta}),$$
where $\theta$ is chosen so that $F$ also has an essential singularity. (There is at most one $\theta$ where this is not the case.) Then there exist points tending to infinity for which $F(z)$ is arbitrarily close to $1$, which means that the points $w_1 := f(z)$ and $w_2 := f(\overline{z} e^{2\pi i \theta})$ have the property that $w_1\cdot\overline{w_2}$ is close to $1$. This means that the two points are essentially on opposite points of the Riemann sphere. So if we map a circle $C$ centered at zero and passing through $z$ under $f$, the image must have length at least $\pi$ (up to a small error, tending to zero). Of course $C$ itself has length $2\pi |z|$, so there must be a point where the spherical derivative is at least $|z|/2$ (again up to an error tending to zero). 
A: These are some quick thoughts, but only give a partial answer. Apologies if I am missing something!
If I understand correctly, you are assuming that f is a meromorphic function in the complex plane.
(Or perhaps you do mean entire holomorphic functions, as you state, in which case you should ask whether $f$ is a polynomial.)
Note that the quantity you are writing down is the spherical derivative, i.e. the derivative measured with respect to the Euclidean metric in the domain and the spherical metric in the range.
Let us assume for a moment that $f$ satisfies the stronger property that
   $$ \frac{|f'(z)|}{1+|f(z)|^2} = o\left(\frac{1}{z}\right).$$
Consider the image of any large circle around the origin under $f$. Then its image under $f$ has spherical length tending to zero. You should be able to show that this means the function extends continuously to infinity, and hence is rational.
With your stronger condition, the image of the curve described stays bounded, but no longer has to tend to zero. This seems like an odd property for a transcendental meromorphic function, but I do not see a simple argument that the function would have to be rational in this case (or a simple counterexample). A counterexample would have to have very small growth in the sense of Nevanlinna theory, so perhaps Nevanlinna theory could help you to investigate this further. I expect that experts in the field would know the answer, so perhaps there is one here to comment?
