$\DeclareMathOperator{\ass}{Ass}$
Let $R$ be a Noetherian ring and $S$ be a multiplicative set in $R$. Then show that $\ass_R(S^{-1}M)=\ass_R(M)\cap \{P\in \text{Spec}(R):P\cap S=\emptyset\}$
Attempt: Let $P\in \ass_R(S^{-1}M)$. We argue that $S\cap P=\emptyset$. Say $P=\text{Ann}(m/s)$ for some $m/s\neq 0$ in $S^{-1}M$. If $r\in P\cap S$, then from $r\cdot (m/s)=0$ we get $(r/1)\cdot (m/s)=0$. But $r/1$ is invertible in $S^{-1}R$ and thus we have $m/s=0$, a contradiction. So we must have $S\cap P=\emptyset$.
The main difficulty I am having is in showing that $P\in \ass_R(M)$. I suspect that $P=\text{Ann}(m)$. Clearly, if $r\cdot m=0$, then $r\cdot (m/1)=0$ and therefore $r\in P$. So $\text{Ann}(m)\subseteq P$. On the other hand, if $r\in P$, then $(r\cdot m)/s=0$, giving an element $u\in S$ such that $(ur)\cdot m=0$. But I am getting nowhere with this.
$\mathfrak p\in\operatorname{Ass}_{R}(S^{-1}M)\implies\exists x\in M,\ \mathfrak p=\operatorname{Ann}_{R}(x/1)$.
This shows that $\mathfrak p\supseteq\operatorname{Ann}_R(x)$. (If you think at this point that $\mathfrak p=\operatorname{Ann}_R(x)$, forget it!)
Moreover, $\mathfrak p$ is minimal over $\operatorname{Ann}_R(x)$: if $\mathfrak p\supseteq\mathfrak p'\supseteq\operatorname{Ann}_R(x)$, then let $a\in\mathfrak p$, and from $\frac a1\cdot\frac x1=\frac01$ deduce that there is $s\in S$ such that $sa\in\operatorname{Ann}_R(x)$, so $sa\in\mathfrak p'$ hence $a\in\mathfrak p'$.
Now one uses that $R$ is Noetherian and conclude that $\mathfrak p\in\operatorname{Ass}_{R}(M)$.