Let $P$ be a maximal ideal of a noetherian ring $R$ and let $P^{-1}=\{x\in Frac(R)|xP\subset R \}.$
I want to show that $R\neq P^{-1}$.
Let $\alpha\neq 0$ in $P$ and choose $r$ minimal so $(\alpha)$ contains the product of $r$ prime ideals $q_1...q_r\subset (\alpha)\subset P$ (Becasue $R$ is noetherian). One of the $q_i$ is contained in $P$ since $P$ is prime. Let $q_1$ be that ideal.
My goal is to show that $q_1$ is maximal.
This is false (assuming that your question makes sense by adding the assumption that $R$ is a domain). Take $R=\mathbb{Z}[X]$ and $P=(2,X)$. Then $R$ is a Noetherian domain and $P$ is maximal, but $P^{-1}=R$.
Indeed, we have $P^{-1}=\{ f\in\mathbb{Q}(X)\mid 2f, Xf\in\mathbb{Z}[X]\}$.
Write $f=\dfrac{g}{h}$ with $g,h\in\mathbb{Z}[X]$ coprime. If $f\in P^{-1}$, $h\mid 2g$ and $h\mid Xg$. Since $g,h$ are coprime and $\mathbb{Z}[X]$ is a UFD, Gauss lemma shows that $h\mid 2$ and $h\mid X$. One easily deduce that $h=\pm 1$ , since $2,X$ are coprime in $\mathbb{Z}[X]$. Therefore, $f=\pm g\in\mathbb{Z}[X]$, and we get $P^{-1}=\mathbb{Z}[X]=R.$
It is true however if $R$ is a Dedekind domain. if $P=0$, then $P^{-1}=Frac(R)\neq R$
Assume that $P\neq 0$. It is enough to show that $R_P\neq (P^{-1})_P$. Since $Frac(R_P)=Frac(R)$, we have $(P^{-1})_P=(P_P)^{-1}$, and one may assume that $R$ is a local Dedekind domain, with maximal ideal $P$. But a local Dedekind domain is a PID. Writing $P=(\pi)$, with $\pi$ irreducible (since $P$ is prime), we see that $\dfrac{1}{\pi}\in P^{-1}$. Now, $\dfrac{1}{\pi}\notin R$ since $\pi$ is not a unit.