= Solution
Let $J$ be an <ideal> of $R'=S^{-1}R$ and contract it to
$$
I=\{r\in R:r/1\in J\}.
$$
This is an ideal of $R$, so $I=(d)$ because $R$ is a <principal ideal domain>. Every $x^n/1$ is a <unit in a ring> $R'$. Consequently, if $r/x^n\in J$, then $r/1=(x^n/1)(r/x^n)\in J$, so $r\in(d)$ and $r/x^n\in(d/1)$. Conversely $d/1\in J$, hence every multiple of it belongs to $J$. Thus
$$
\boxed{J=(d/1),}
$$
so every ideal of $R'$ is principal. The localization remains an <integral domain>; hence \b[$R'$ is a principal ideal domain.]
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