Let $k$ be an algebraically closed field.
Suppose we have a discrete valuation subring $A$ of $k(x)$ containing $k$ and whose quotient field is exactly $Frac(A)=k(x)$. If we are given that $x\not\in A$, then I have to prove that $A=\left \{\frac{f}{g}\in k(x):\deg(f)\leq \deg(g)\right \}$.
We know, of course, that $\left \{\frac{f}{g}\in k(x):\deg(f)\leq \deg(g)\right \}$ is a discrete valuation subring of $k(x)$ satisfying all the given properties for $A$. Its maximal ideal is the ideal generated by $x^{-1}$.
Let $m$ be the maximal ideal of $A$. Since $x\not\in A$, then $x^{-1}\in m$. Therefore, all we have to do is to prove that $\left \{\frac{f}{g}\in k(x):\deg(f)\leq \deg(g)\right \}\subseteq A$. If we were able to do so, then the maximal ideal of the first ring would be contained in $m$, because it is generated by $x^{-1}\in m$. And then we would be able to make use of the following fact:
Let $A,B$ be two discrete valuation rings such that $A\subseteq B\subseteq Frac(A)$. Let $m$ be the maximal ideal of $A$ and let $n$ be the maximal ideal of $B$. If $m\subseteq n$, then $A=B$.
By using this fact, we would conclude that $A=\left \{\frac{f}{g}\in k(x):\deg(f)\leq \deg(g)\right \}$, which is exactly what we are trying to achieve.
So, we would be done if we were able to prove that $\left \{\frac{f}{g}\in k(x):\deg(f)\leq \deg(g)\right \}\subseteq A$. I am not succeeding in proving it. I tried by supposing that there exists an $f/g$ in the former set which is not in $A$ and conclude that $g/f\in m$, but I am not attaining any contradiction from this.
How would you proceed?
Put $R = \{f/g \in k(x) : \deg(f) \leq \deg(g)\} \subset k(x)$. First, let us recall how one shows that $R$ is a local ring with principal maximal ideal, generated by $x^{-1}$.
Indeed, since $\deg(f) = \deg(g)$ is (manifestly) a symmetric condition, we see that the subset of elements $S := \{f/g \in k(x)^{\times} : \deg(f) = \deg(g)\} \subset R$ are all units, and every element in the complement of $S$ is not invertible for straightforward degree reasons, and so $R^{\times} = S$. Meanwhile, $M := R \setminus S = \{f/g \in k(x) : \deg(f) < \deg(g)\}$ is easily seen to be an ideal of $R$, and so we conclude that $R$ is local, and $M$ is the unique maximal ideal of $R$. Clearly, $1/x \in M$. Conversely, Let $f/g \in M$ be nonzero. Since $\deg(f) < \deg(g)$, there exists a natural number $n > 0$ such that $\deg(x^{n}f) = \deg(g)$. In particular, $f/g = (1/x)^{n}(x^{n}f/g)$, whence $M \subseteq (1/x)R$, and so $M = (1/x)R$.
By carefully going through the proof above, we notice that if $A$ contains $S$, then $A$ contains all of $R$, since every element of $R$ is some element of $S$ scaled by a suitable power of $1/x$. Thus, it suffices to show that $S \subset A$. To do this, let $f/g \in S$. By scaling $f$ and $g$ appropriate by some elements elements of $k$, we may assume that they are both monic. Moreover, since $A$ is a valuation ring of $k(x)$, at least one of $f/g$ or $g/f$ is in $A$. WLOG, assume that $f/g$ is in $A$; we will show that $g/f$ is also in $A$, i.e. that $f/g$ belongs to $A^{\times}$. This will complete the proof, since it shows that every element of $S$ belongs to $A$. Since $A$ is local, it suffices to show that while $f/g$ is in $A$, $f/g$ does not belong to the unique maximal ideal $M'$ of $A$.
Towards contradiction, suppose $f/g \in M'$. Put $n = \deg(f) = \deg(g) > 0$. Now, let $h(x) = g(x)-f(x)$; since we have assumed that $f$ and $g$ are monic, note that $\deg(h) < n$. Put $f(x) = x^{n}+r(x)$, Note that $$(1/x)(f(x)/g(x)) = x^{n-1}/g(x) + (1/x)(r(x)/g(x)).$$
I claim that $r(x)/g(x) \in A$. If $r(x) = 0$, then we're done. If $r(x) \neq 0$, then either $r(x)/g(x) \in A$ or $g(x)/r(x)$ is in $A$. If the latter is the case, then $g(x)/r(x) \cdot f(x)/g(x) = f(x)/r(x)$ is in $A$. Since $f(x) = x^{n}+r(x)$, this would imply that $x^{n}/r(x)$ is in $A$, and so $x^{i}/r(x)$ is in $A$ for all $0 \leq i \leq n$. In particular, since $d := \deg(r(x)) < n$, we see that $(x^{n-d}r(x))/r(x) = x^{n-d}$ is in $A$, a contradiction.
Hence, we see that $(1/x)(r(x)/g(x)) \in M'$, so $x^{n-1}/g(x)$ belongs to $M'$. Scaling by powers of $1/x$, we then obtain that $x^{i}/g(x)$ belongs to $M'$ for all $0 \leq i \leq n-1$. In particular, we see that $h(x)/g(x) \in M'$, and so $f(x)/g(x) + h(x)/g(x) = g(x)/g(x) = 1 \in M'$, a contradiction. This completes the proof.