If $M$ is a maximal ideal of a commutative ring R, then $R/M \cong R_M/M_M$, where $R_M$ is the localization of $R$ at $M$.
We know that localization commutes with taking quotients, so $R_M/M_M = (R/M)_M$. If $a \in R/M$, then we obviously have $a \in (R/M)_M$.
Now let $a/b \in (R/M)_M$. We know that $a \in R/M$. Also, since $b \not\in M$ and since $R/M$ is a field, $1/b \in R/M$. So $a/b \in R/M$. Then we get $R/M = (R/M)_M = R_M/M_M$. So I'm not really sure why the question says "isomorphic" and not "equal". Did I miss something?
Both equalities are just shorthand for "isomorphic". So, when we write $(R/M)_M =R_M/M_M$ we actually mean $(R/M)_M \cong R_M/M_M$.
Formally, this is a dangerous notation because there is a difference between $\cong$ and $=$. However, similar to the common joke stating that a topologist can't tell a mug from a doughnut, an algebraist is often only interested in the module structure of the considered object. If the structure is the same due to a canonical isomorphism, one sometimes writes $=$ instead of $\cong$.
In your case, $(R/M)_M \cong R_M/M_M$ is canonically given by $(a+M)/b \mapsto a/b +M_m$ for $a \in R$,$ b \in R\setminus M$. However, the elements of $(R/M)_M$ are not physically contained in $R_M/M_M$ meaning that $(R/M)_M=R_M/M_M$ does not hold in the set theory interpretation of the symbols.