Degree of an holomorphic map and ramification points how to determine deg f
$f(z)=\frac {z^{3}}{1-z^{2}} $
and what are its ramification points
The ramification point i found $p \in \mathbb{C}^{\infty}$ is a point with $multp(F) \geq 2$.
I found for example the following point:
$p=0$ because its a zero of f and so $multp(F)=ordp(F)=3 \geq 2$
The poles $+1$ and $−1$ are no ramification points sich their order are just $−1$. Also $\infty$ is no ramification points since
$f(\frac{1}{z}) = \frac{\frac{1}{z^3}}{1-\frac{1}{z^2}} = \frac{\frac{z^2}{z^2-1}}{z^3} =  \frac{1}{z(z^2-1)}$
has a pole of order just 1 in zero (ord1(F)=−1).
 A: The map $f\colon\mathbb{C}_\infty\rightarrow\mathbb{C}_\infty; z \mapsto \frac{z^3}{1-z^2}$ on the Riemann sphere $\mathbb{C}_\infty$ has degree $3$. Its ramification points are $\pm\sqrt{3}$ and $0$. The points $\pm\sqrt{3}$ each have ramification index $2$, while $0$ has ramification index $3$.
Let $p\in\mathbb{C}$. Assume that $p$ is not a pole of $f$. By Lemma 4.4 in Miranda's Algebraic Curves and Riemann Surfaces the multiplicity of $f$ at $p$ is given by the exponent of the first strictly positive term in the Taylor expansion of $f$ around $p$. Thus, $p$ is ramification point of $f$ if and only if the first derivative $f'(z)=\frac{z^2(3-z^2)}{(1-z^2)^2}$ of $f$  vanishes at $p$. Now, the zeroes of $f'$ are precisely $\pm\sqrt{3}$ and $0$. Additionally, by considering higher order derivatives, we can read off the ramification indices.
In the next step, take the point $\infty\in \mathbb{C}_\infty$. Since $f(\infty)=\infty$, we have to consider the Taylor expansion of $\frac{1-(\frac{1}{z})^2}{(\frac{1}{z})^3}=z^3-z$ around $0$. The exponent of the first strictly positive term of that Taylor expansion is $1$. Hence, $f$ has multiplicity $1$ at $\infty$. Consequently, $\infty$ is not a ramification point.
Now, let $p\in \mathbb{C}$ be a pole of $f$, i.e. $p=\pm 1$. Then
$\operatorname{mult}_p(f)=–\operatorname{ord}_p(f)=1$, since $p$ is a pole of order $1$. Thus, in this case, $p$ is not a ramification point.
Finally, since $f^{-1}(0)=\{0\}$ holds, we can calculate the degree of the proper map $f$ as follows:
$$\operatorname{deg}(f)=\sum_{x\in f^{-1}(0)}\operatorname{mult}_x(f)=\operatorname{mult}_0(f)=\operatorname{ord}_0(f)=3.$$

Alternatively, the degree of $f$ can be read off directly by employing the following result (and noting that the degree of a nonconstant polynomial is precisely its degree as a proper nonconstant holomorphic mapping).
Lemma. Let $g$ and $h$ be two coprime univariate complex polynomials. Assume that $g$ is nonconstant. Then the degree $d$ of the rational mapping $f=\frac{g}{h}\colon \mathbb{C}_\infty\rightarrow\mathbb{C}_\infty$ is given by $d=\operatorname{max}\{\operatorname{deg}(g),\operatorname{deg}(h)\}$.
Proof. First assume that $\operatorname{deg}(h)\geq \operatorname{deg}(g)$. Denote the set of zeroes of $h$ by $Z$. Since $\operatorname{deg}(h)\geq \operatorname{deg}(h)$, we have $Z=f^{-1}(\infty)$. Thus
$$\operatorname{deg}(f)=\sum_{x\in f^{-1}(\infty)}\operatorname{mult}_x\big(\frac{g}{h}\big)=\sum_{x\in Z}\operatorname{mult}_x\big(\frac{g}{h}\big)=\sum_{x\in Z}-\operatorname{ord}_x\big(\frac{g}{h}\big)= \sum_{x\in Z}\operatorname{ord}_x(h)-\sum_{x\in Z}\operatorname{ord}_x(g).$$ Since $g$ and $h$ are corprime, we have $\sum_{x\in Z}\operatorname{ord}_x(g)=0$ and therefore $\operatorname{deg}(f)=\operatorname{deg}(h).$
The case $\operatorname{deg}(g)\geq \operatorname{deg}(h)$ is treated analogously by looking at the point-preimage $f^{-1}(0)$.

Alternatively, some of the quantities that you asked for can be determined by the Riemann-Hurwitz formula. Suppose, for instance, that you calculated the degree of $f$ as $3$ and found all three ramification points. The multiplicity of $f$ at $0$ is just the order of the zero, namely $3$. Riemann-Hurwitz then implies $2=\sum_{p\in\{\pm\sqrt{3}\}}\operatorname{mult}_p(f)-1$. Hence, the ramification index at the points $\sqrt{3}$ and $-\sqrt{3}$ is $2$ in each case.
