How to show there are only two discrete valuation rings with quotient field $k(x)$? I want to show that only discrete valuation rings with quotient field as $k(x)$ containing $k$ are: 
$\mathcal{O_{a} (\mathbb{A^{1}})}$ for each $a \in k$ and $\mathcal{O_{\infty}}$;
the former is the set of rational functions on $\mathbb{A^{1}}$ (affine 1-space, that is field $k$ here) that are defined at $a \in k$, it is a discrete valuation ring with uniformizing parameter $x-a$ and the latter is the ring
$$
\left\{\frac{F}{G} \in k(x) \mid \deg(G) \geq \deg(F) \right\}
$$
with $\frac{1}{x}$ as its uniformizing parameter.
My idea was to first observe that if $S$ is any DVR, then it cannot be clearly field of quotients $k(x)$, since in the book (Fulton, Algebraic Curves) we have not defined them as fields.
So, $S\subset k(x)$.
It will contain the ring $k[x]$. Now I will use a previous exercise that says that 
 "If $R$ is a DVR with quotient field $K$ and $m$ as its maximal ideal then for $z\in K, z \notin R$, we must have $z^{-1} \in m$."
and another that says that 
"Further if $R\subset S\subset K$ and $S $ is also a DVR, and the maximal ideal of $S$ contains $m$ then $S =R$."
But I don't know how I can start.
Any hint would be appreciated, thanks! 

Let me start by pointing out the a DVR $R\subset k(x)$ may not contain $k[x]$, and in fact you $\mathscr{O}_\infty$ doesn't. Here's an outline of how to prove the result: (1) using the second exercise you mention (about maximality of DVRs), prove that a DVR R that does contain $k[x]$ must be the localization of $k[x]$ at some maximal ideal, and so isomorphic to some $\mathscr{O}_a (\mathbb{A}^1)$ (hint: if $\mathfrak{m}$ is the maximal ideal of $R$, what is $\mathfrak{m}\cap k[x]$?) (2) by the first exercise you cite, if $R$ does not contain $k[x]$, it must contain $x^{-1}$; use maximality again to prove that such a DVR must be $\mathscr{O}_\infty$.
