Automorphisms of $\mathbb P^1_{\mathbb k}$ over a algebraically closed field $\mathbb k$. 
Let's consider an algebraically closed field $\mathbb k$. Consider the birational map $\phi: \mathbb P^1_{\mathbb k} \to \mathbb P^1_{\mathbb k}$. How do I show that if there are $f,g$, homogeneous poynomials, such that $$\phi[x:y] = [f(x,y):g(x,y)],$$  then $\deg(f) = \deg(g)=1$? In other words, how do I show that $\phi$ is a element of $\mathrm{PGL}_2(\mathbb k)$?

So far I was able to prove that $\deg(f) = \deg(g)$. Let me give you this part of the argument: Take $[x:y]\in\mathbb P^1_{\mathbb k}$ such that $\phi[x:y]=[1:1]$. We must have $$ f(\mu x,\mu y)g(x,y) -g(\mu x,\mu y)f(x,y)=0\quad \forall  \mu \in \mathbb k\setminus\{0\},$$
because $[f(x,y):g(x,y)]=[f(\mu x,\mu y):g(\mu x,\mu y)]$. Using that $f(x,y)=g(x,y)\neq 0$ we conclued that the above equation can be written as
$$ [\mu^{\deg(f)}-\mu^{\deg(g)}]f(x,y)^2=0\quad \forall  \mu \in \mathbb k\setminus\{0\}$$
and therefore $\deg(f)=\deg(g)$.

Let me just summarize what was discussed on the comment section. 
A birational map $\phi$ is given by a rational function $z \to \frac{f(z)}{g(z)}$ with $f,g$ polynomials. The inverse is also given by a rational function. By this fact we obtain that $z \to \frac{f(z)}{g(z)}$ is automorphism of $\mathbb k(z)$ and the automorphisms of k(z) are given by the Mobius transformations. Therefore $$\frac{f(z)}{g(z)} = \frac{az+b}{cz+d}$$
where $ad-bc \neq 0$.
If $\phi$ is rational and bijective (this was my original hypothesis) then $\phi$ is not necessarily given by Mobius transformations. The counter-example is given by $\phi[x:y] = [x^p:y^p]$ over a  field $\mathbb k$ of characteristic $p$.
