Let $R$ be a commutative ring, $M$ is a maximal ideal in $R$, $P$ is a prime ideal and $M\supset P$. I want to prove that $R_{P}$ is a localization of $R_{M}$.
Intuitively, I would say $R_{P}$ is a localization of $R_{M}$ at $P_M$. For if $s_1,s_2 \notin M$, $r\in R$ and $p^*\notin P$ we have $ (r_1/s_1)/(p^*/s_2)=(r_1s_2)/(s_1p^*)$ and $s_1p^*\notin P$.
Conversely, let $r/p^* \in R_P$ we have $r/p^*=(r/1)/(p^*/1)$ hence $R_{P}$ is a localization of $R_{M}$.
This argument seems to be informal so I'm looking for rigorous proof.
Prime ideals of $R_M$ naturally correspond to prime ideals of $R$ contained in $M$. So in particular, $P$ naturally corresponds to a prime ideal of $R_M$ (precisely, if we let $S=R\smallsetminus M$, then the corresponding prime ideal is $S^{-1}P$). What you want to show is that there is a "natural" isomorphism between $R_P$ and $(R_M)_{S^{-1}P}$.
One way to do this is to show that $(R_M)_{S^{-1}P}$ satisfies the universal property of localization for $R_P$:
From this universal property definition, you it follows that $S^{-1}R$ is unique "up to unique isomorphism" (whatever that means, it doesn't matter here).
With iterated use of the universal property on $R_M$ then on $(R_M)_{S^{-1}P}$, you should be able to show it in fact this ring satisfies the universal property for $R_P$, and hence $(R_M)_{S^{-1}P}\cong R_P$.
On the other hand, you can try to go the "naive" route to show the isomorphism, which will work fine but is more tedious. What do elements of $(R_M)_{S^{-1}P}$ look like? They are elements $b/t$ with $b\in R_M$ and $t\in S^{-1}P$, so rather, they are elements of the form $(a/s)/(b/t)$ where $a\in R,s,\in S$, and $b/t\notin S^{-1}P$, which is the same as saying that $t\in S$ and $b\notin P$. Then you can define a morphism $(R_M)_{S^{-1}P}\to R_P$ by sending $(a/s)/(b/t)\mapsto (at)/(bs)$. Note that indeed since $s\notin M$, $s\notin P$ as well, so $bs\notin P$. You can check for yourself that this is well-defined, a homomorphism, bijective, etc.