$F[x{]}_{(x)}$ is the ring of rational functions defined at $0$, with maximal ideal generated by $x$ and residue field $F$

By [F1] and [F2], evaluation at $0$ has kernel $(x)$ and is onto. Explicitly, $f+(x)\mapsto f(0)$ is well defined and has inverse $c\mapsto c+(x)$, so $F[x]\mathrm{/}(x)\cong F$. Thus [F3] makes $(x)$ maximal, and [F6] makes it prime.

By [F9], the denominators are the elements outside $(x)$, which are exactly the polynomials $g$ with $g(0)\mathrm{\ne }0$ by [F2]. They are nonzero and hence units in $F(x)$ by [F5], so [F7] gives a map from the localisation to $F(x)$ sending $f\mathrm{/}g$ to the same fraction. If its image is zero, then $f=0$ because [F5] embeds the domain $F[x]$ in its fraction field, and [F8] makes $f\mathrm{/}g=0$ in the localisation. The map is therefore injective, and its image is precisely the displayed fractions, which are exactly those admitting a representative with denominator nonzero at $0$.

By [F4], the maximal ideal is $(x)F[x{]}_{(x)}=xF[x{]}_{(x)}$, and the residue field is $\mathrm{Frac}(F[x]\mathrm{/}(x))\cong \mathrm{Frac}(F)$. Every nonzero denominator in the field $F$ is already invertible, so $a\mathrm{/}b=(a{b}^{-1})\mathrm{/}1$ and the last fraction field is canonically $F$.