Determine all discrete valuations on $\mathbb{C}(x)$. To clarify, for a field $K$, a valuation $v$ on $K$ is a map $v:K^{\times}\to G$ for $G$ an ordered group (written additively) such that for any $a,b\in K^{\times}$:
1) $v(ab)=v(a)+v(b)$;
2) $v(a+b)\geq\min\{v(a),v(b)\}$ if $a+b\neq 0$.
We extend $v$ to $K$ by setting $v(0)=\infty$. A valuation is a discrete valuation if $v(K^{\times})$ is $\mathbb{Z}$.
There is a hint which says we could start by determining all the valuation rings. But I do not know how to use this. So I started from the basic by determining $v(c)$ for $c\in\mathbb{C}^{\times}$ and $v(x)$--which then determines $v$.
If we assume $v(c)=0$ for all $c\in\mathbb{C}^{\times}$, then:
For $v(x)$, we consider the cases $v(x)<0$ and $v(x)\geq 0$--that is, $v(x)\in\mathcal{O}_v$ or $v(x)\notin\mathcal{O}_v$.
Case $v(x)<0$: Then $v(x^{-1})>0$ and for any $f\in\mathbb{C}[x]$ of degree $d$, $v(f)=dv(x)=-dv(-x)$. Then for $q=f/g\in\mathbb{C}(x)$, $v(q)=(-deg(f)+deg(g))\cdot v(-x)$. So it seems to me that in this case, if we require $v$ to be a discrete valuation, $v(\mathbb{C}(x))$ is infinitely cyclic with $v(x)$ be its generator. Namely, we get a discrete valuation which sends $q=f/g$ to $v(q)=-deg(f)+deg(g)$.
Case $v(x)\geq 0$: Then $\mathbb{C}[x]\subset \mathcal{O}_v$. Consider $I=\mathbb{C}[x]\cap\mathcal{m}_v$. Notice that $I\neq(0)$. Since $m_v$ is the maximal ideal of $\mathcal{O}_v$, $I$ is a prime ideal of $\mathbb{C}[x]$, which is a PID. Then I is maximal. Then $I=(x-\alpha)$ for some $\alpha\in\mathbb{C}$. For any $r\in\mathbb{C}(x)$, write $r=(x-\alpha)^m\cdot\frac{f}{g}$ such that $f,g\in\mathbb{C}[x]\setminus(p)$. Then $v(r)=v((x-\alpha)^m)$ since $v(f)=v(g)=0$. Then $v(r)=mv(x-\alpha)$. If we require $v$ to be discrete, then $v(x-\alpha)=1$. 
We are done with the case for discrete valuations that are trivial on $\mathbb{C}^{\times}$.
What happens when the $v$ is not necessarily trivial on $\mathbb{C}^{\times}$?
Is it possible to find all valuation ring of $\mathbb{C}(x)$ first and then pick the discrete ones so that we can recover the valuation?
 A: Here's a guide to finding the discrete valuations on $\mathbb{C}(x)$. I've put my solutions in spoiler boxes so you can try before you peek. Also, since you didn't mention it, $A$ is a valuation ring of $\mathbb{C}(x)$ iff it is a subring and $z\in A$ or $z^{-1}\in A$ for all non-zero $z\in\mathbb{C}(x)$. In particular, valuations rings are local.
The main idea is that for a discrete valuation ring $A$ with maximal ideal $\mathfrak{m}$, the valuation $v(x)$ is the greatest integer $n$ such that $x\in\mathfrak{m}^n$.
Suppose that $A$ is a valuation ring of $\mathbb{C}(x)$.
First, show that $\mathbb{C}\subseteq A$, if $A\cap\mathbb{C}$ is assumed to be a closed subset of $\mathbb{C}$. Without this assumption you may run into some set theoretic issues or non-discrete valuations.

 For every $\zeta\in\mathbb{C}$ on the unit circle, $\zeta\in A$ or $\zeta^{-1}\in A$. Thus $A$ contains some $\zeta$ with argument an irrational multiple of $\pi$. Since $A\cap\mathbb{C}$ is closed and the orbit of $\zeta$ is dense, $A$ contains $S^1$. Since $A$ is a ring, $A$ contains $\mathbb{Z}$. The set $S^1\cdot(\mathbb{Z}+S^1)$ (i.e. integer translates of the circle followed by arbitrary rotations) yields all of $\mathbb{C}$. So $\mathbb{C}\subseteq A$.

Note that $\mathbb{C}(x)$ is a valuation ring of itself. What is the valuation in this case?

 $G=\mathbb{C}(x)^{\times}/\mathbb{C}(x)^{\times}$ is an ordered (trivial) group. The quotient map $v:\mathbb{C}(x)^{\times}\rightarrow\mathbb{C}(x)^{\times}/\mathbb{C}(x)^{\times}=G$ has the properties that $v(ab)=v(a)+v(b)$ and $v(a-b)\geq\min(v(a),v(b))$, if $a\neq b$ (where we are writing $G$ additively). So we have obtained the trivial valuation.

Now assume that $A$ is a proper subring. Since $A$ is a valuation ring of $\mathbb{C}(x)$, $x\in A$ or $x^{-1}\in A$.
Find an explicit description of $A$ assuming that $x\in A$.

 Since $x\in A$, $\mathbb{C}[x]$ is a subring of $A$. Since $A$ is a valuation ring, it is local. Let $\mathfrak{m}$ be the unique maximal ideal of $A$. The preimage of $\mathfrak{m}$ under the inclusion $j:\mathbb{C}[x]\rightarrow A$ is a prime ideal of $\mathbb{C}[x]$. Since $\mathbb{C}$ is algebraically closed, $j^{-1}(\mathfrak{m})=(x-a)$ for some $a\in\mathbb{C}$. Thus every element of $\mathbb{C}[x]$ not in $(x-a)\mathbb{C}[x]$ is invertible in $A$. By the universal property of localization, $j$ lifts to some map $\overline{j}:\mathbb{C}[x]_{(x-a)}\rightarrow A$. Furthermore, $\overline{j}$ is injective because $\mathbb{C}[x]$ is an integral domain. The only elements of $\mathbb{C}(x)$ not in $\mathbb{C}[x]_{(x-a)}$ are those with a factor of $(x-a)$ in the denominator. Thus $\mathbb{C}[x]_{(x-a)}$ is a maximal proper subring of $\mathbb{C}(x)$. Since $A$ was assumed to be proper, $\overline{j}$ is surjective, hence an isomorphism.

Derive what the valuation must be for this description of $A$.

 The ring $\mathbb{C}[x]_{(x-a)}$, for any $a\in\mathbb{C}$, is easily seen to be a valuation ring of $\mathbb{C}(x)$. The valuation is $v:\mathbb{C}(x)^{\times}\rightarrow\mathbb{C}(x)^{\times}/\mathbb{C}[x]_{(x-a)}^{\times}=G$. Note that the elements of $G$ are uniquely identified by coset representatives of the form $(x-a)^n$, where $n\in\mathbb{Z}$. So $G$ is isomorphic to $\mathbb{Z}$. Choose the identification of $(x-a)\in G$ with $1\in \mathbb{Z}$ (there are two possible isomorphisms, this picks one). The ordering on $G$ is just the ordering of the integers (powers of $(x-a)$). Identifying $G$ and $\mathbb{Z}$, we get $$v((x-a)^mp(x)+(x-a)^nq(x))=v((x-a)^k[(x-a)^{m-k}p(x)+(x-a)^{n-k}q(x)])\geq v((x-a)^k)=k,$$ where $p(x),q(x)$ are assumed to have no factors of $(x-a)$ in numerator or denominator, and $k=\min(m,n)$, $m,n\in\mathbb{Z}$. Similarly, $$v((x-a)^mp(x)\cdot(x-a)^nq(x))=v((x-a)^{m+n}p(x)q(x))=m+n.$$ So $v$ is in fact a discrete valuation of $\mathbb{C}(x)$.

Now assume that $x^{-1}\in A$ and $x\notin A$. Find a valuation ring and its valuation that were not found above.

 We also have that the ring $\mathbb{C}[x^{-1}]_{(x^{-1})}$ is a valuation ring of $\mathbb{C}(x)$. One way to see this is that $x\mapsto x^{-1}$ is an automorphism of $\mathbb{C}(x)$, and $\mathbb{C}[x]_{(x)}$ is one of the valuation rings from the previous paragraph. Repeating the same construction as before yields a discrete valuation.

Finally, show that there are no other discrete valuation rings of $\mathbb{C}(x)$.

 Our characterization in the case that $x\in A$ was forced. So there cannot be any others in that case. Suppose that $x\notin A$. If we run through the characterization of $A$ analogously to the case $x\in A$ we end up with $A=\mathbb{C}[x^{-1}]_{(x^{-1}-a)}$ for some $a\in\mathbb{C}$. If $a\neq0$, then $x^{-1}$ is a unit of $A$ because it is not in the maximal ideal of $A$. This is a contradiction, so we are forced to have $a=0$ and we end up with the valuation ring from before. Therefore, the only wiggle room on valuations and valuation rings of $\mathbb{C}(x)$ comes from dropping the assumption that $A\cap\mathbb{C}$ is a closed subset of $\mathbb{C}$. This will lead to considerations of possibly non-measurable subsets of $\mathbb{C}$ and other set theoretic difficulties that I cannot at present address. Furthermore, it should not yield any more discrete valuations, but I cannot prove that right now either.

