In which condition: an element of function field of $\mathbb{P}^1$ has zero or pole or no-zero&no-pole.
I am thinking that: since $\mathbb{P}^1$ and $\mathbb{A}^1$ is birrationally equivalent (i.e. their function fields are isomorphic), we can check element in function field of $\mathbb{A}^1$ and (using with homogenize and dehomogenize) we can conclude the result also for $\mathbb{P}^1$.
I actually also wonder that how can we prove "birrationally equivalent iff their function fields are isomorphic." I proved it as in the book of Hartshore. But if we write maps as $[\phi_1,\phi_2,\phi_3]$, then how can we prove in this way.
If $M$ is an (irreducible) variety, then the rational functions are the regular functions, defined on some (non-empty) open subsets (different for different functions), up to the equivalence on the intersection of the subsets. It is well defined because any two (non-empty) open subsets are intersecting.
One may restrict any rational function on the fixed affine open subset $\operatorname{Spec} A$, getting an isomorphism $$\operatorname{Quot} A=k(\operatorname{Spec} A)=k(X),$$ which for $\mathbb A^n \subset \mathbb P^n$ is $$k(x_1,\ldots,x_n)=k(\mathbb A^n)=k(\mathbb P^n),$$ so the rational functions on $\mathbb P^n$ are the "real" rational functions.
In the case of $\mathbb P^1$, any rational function $f \in k(x)$ may be written as a fraction of two coprime polynomials $f=g/h, g,h \in k[x]$. Then the zeroes of $f$ on the map $\mathbb A^1$ are the zeroes of $g$, and the poles are the zeroes of $h$, with the necessary multiplicities. The order of a zero of a pole at the infinity is equal to $\deg f=\deg h-\deg g$, as you can easily check in another map.
So the number of poles is equal to the number of zeroes (counted with multiplicities), and a function either has a zero and a pole, or none of them.