Proof that rational functions are an ordered field, but non-archimedean - Bartle's elements of real analysis
Consider the following order: let $f(x)=p(x)/q(x)$ be a rational function with
\begin{aligned}p(x)=a\cdot x^n &+ \text{ terms of degree less than $n$}\\ q(x)=b\cdot x^m &+ \text{ terms of degree less than $m$}\end{aligned}
where of course $a,b \in \mathbb{Q}$. Our order says that $f > 0$ if and only if $\frac{a}{b} >0$. Notice this defines the order throughout the field; if one wishes to determine whether $f_1 > f_2$, write the difference $f_1-f_2$ as a single rational function and determine whether it is $>0$, $=0$ or $<0$.
Now, this totally ordered field is not Archimedean. Indeed, consider the rational functions $f(x)=x$ and $g(x) = 1$. No matter how large you choose $n \in \mathbb{N}$, $f(x)>n\cdot g(x)$, because $\big(f-n\cdot g\big)(x)=x-n$ and the leading coefficient of $\big(f-n\cdot g\big)$ is $1$, which is positive.
The order is "eventual domination": $f(x)\geq g(x)$ iff for all sufficiently large $q\in\mathbb{Q}$, $f(q)\geq g(q)$. It takes a bit of work to show that this really is a total order on rational functions. For a more explicit version of this definition, if $f(x)=\frac{ax^n+\dots}{bx^m+\dots}$ (where the omitted terms have lower degree), then $f(x)\geq 0$ iff $\frac{a}{b}\geq 0$. To determine whether $f(x)\geq g(x)$, you then just write $h(x)=f(x)-g(x)$ in the form $\frac{ax^n+\dots}{bx^m+\dots}$ to determine whether $h(x)\geq 0$.
To see that this order is non-archimedean, just observe that $x>n$ for all $n\in\mathbb{Z}$. Indeed, taking $f(x)=x$ and $g(x)=n$, then $f(q)\geq g(q)$ for all $q\geq n$. In fact, it can be shown that the eventual domination order is the unique ordering on $\mathbb{Q}(x)$ compatible with the field structure for which $x>n$ for all $n\in\mathbb{Z}$. That is, if you declare that $x>n$ for all $n\in\mathbb{Z}$, then the entire rest of the ordering can be deduced from the ordered field axioms. The intuition is that if $x$ is infinitely large, then $f(x)$ behaves like $f(q)$ for very large $q$, so you can determine whether $f(x)\geq g(x)$ by comparing $f(q)$ and $g(q)$ for large $q$.
For polynomials $f(x),g(x)\in\mathbb{Q}[x]$ define $f<g$ if and only if $f\ne g$ and the leading coefficient of $g-f$ is positive.
It's easy to see that this defines a (strict) order relation on $\mathbb{Q}[x]$ compatible with the operations, in the sense that
for every $f\in\mathbb{Q}[x]$ it holds exactly one among $f>0$, $f=0$, or $0>f$;
if $f,g,h\in\mathbb{Q}[x]$ and $f<g$, then $f+h<g+h$
if $f,g,h\in\mathbb{Q}[x]$, $f<g$ and $0<h$, then $fh<gh$.
Now it's easy to see that $0<f$ if and only if $-f<0$, so any element of $\mathbb{Q}(x)$ can be written as a quotient $f(x)/g(x)$ where $0<g$.
Define, for $f_1(x)/g_1(x),f_2(x)/g_2(x)\in\mathbb{Q}(x)$ with $0<g_1$ and $0<g_2$, $$ \frac{f_1(x)}{g_1(x)}<\frac{f_2(x)}{g_2(x)} \quad\text{if and only if}\quad f_1(x)g_2(x)<f_2(x)g_1(x) $$ and prove that this defines a (strict) order relation with the same properties above, so $\mathbb{Q}(x)$ becomes an ordered field.
Now, of $q\in\mathbb{Q}$, it's obvious that $q<x$ and therefore the order on $\mathbb{Q}(x)$ is not Archimedean. Note also that the order induced on $\mathbb{Q}$ is the usual one.
The ordering that I've seen is "$f(x) > 0$ iff $\exists X\in \Bbb R$ s.t. $x>X$ implies $f(x)>0$", in other words, if $f(x) = \frac{p(x)}{q(x)}$, with $p(x) = a_nx^n + \cdots + a_1x + a_0$ and $q(x) = b_mx^m + \cdots + b_1x + b_0$, then $f(x) > 0$ iff $\frac{a_n}{b_m} > 0$.
As usual, this means that $f>g$ iff $f-g > 0$.
It's not archimedean because if $f(x) = x$ and $g(x) = x^2$, then no matter how many times I add together copies of $f$, we will still have that $g$ is greater.