A discrete valuation ring is a domain $R$ that is noetherian, local (i.e., having exactly one maximal ideal) such that its maximal ideal is principal, and $R$ is not a field.
If $R$ is a domain but not a field, then $R$ is a discrete valuation ring if and only if there exists $t\in R$ irreducible such that for every nonzero $r\in R$, $r$ can be written uniquely as $ut^n$ where $n$ is a nonnegative integer and $u$ is a unit.
In that case, $R$ can be shown to be a principal ideal domain and $t$ is the generator of the unique maximal ideal of $R$.
That's all the theory (concerning DVRs) that I am supposed to know in order to solve the following exercise:
Let $k$ be an algebraically closed field. Find all discrete valuation subrings of $k(x)$ that contain $k$ and whose quotient field is $k(x)$.
This is my attempt: since every element of $k(x)$ can be written as $\frac{f}{g}$ where $f,g\in k[x]$, then we can fix $a\in k$ and consider the ring $R_a$ of all rational functions with denominator that does not vanish at $a$. It is obviously a domain that is not a field, so we have to find an adequate $t$. Indeed, $x-a$ is irreducible, for if $x-a=\frac{f}{g}\times \frac{b}{c}$, then $cg$ does not vanish at $a$ and therefore there is exactly one factor of $x-a$ in the factorization of $bf$; so let's assume that $f(a)=0\neq b(a)$. Therefore $\frac{b}{c}$ is a unit in $R_a$ and $x-a$ is irreducible.
Since $x-a$ is irreducible, we can take $t=x-a$ and we have that $R_a$ satisfies the stated equivalence of being a DVR.
I cannot come with other examples of possible solutions. Also, I have not been able to show that these are the only solutions. How would you solve the exercise?