This problem is Exercise 6.4 in Matsumura's Commutative Ring Theory.
Let $I$ and $J$ be ideals of a Noetherian ring $A$. Prove that if $JA_P\subset IA_{P}$ for every $P\in \operatorname{Ass}_A(A/I)$, then $J\subset I$.
The hint in the book is to use a primary decomposition of $I$. (A solution using this hint can be found in this answer.) My proposed solution proceeds differently. Is it correct?
Proposed solution. Note that $J\subset I$ if and only if $\phi(J)=0$, where $\phi:A\rightarrow A/I$ is the quotient map. So we may suppose $I=(0)$. The associated primes $P_i\in\operatorname{Ass}_A(A)$ are those of the form $\operatorname{ann}(x)$ for some $x\in A$.
Suppose $JA_{P_i}=0$ for every $P_i$. Fix $x\in J$. Then $x$ goes to zero in the localization $JA_{P_i}$ if and only if $sx=0$ for some $s\in A\setminus P_i$, if and only if $\operatorname{ann}(x)\not\subset P_i$. We then have $\operatorname{ann}(x)\not\subset P_i$ for every $i$.
Consider the family of ideals $F=\{\operatorname{ann}(y): 0\not= y\in A\} $. Theorem 6.1(i) in Matsumura says that a maximal element in this set is an associated prime. But clearly $\operatorname{ann}(x)$ is contained in some maximal element (this uses the Noetherian hypothesis), so it is contained in some $P_i$, a contradiction.
In fact, I'd take a minimal prime over $\operatorname{Ann}(x)$. Then this is associated, and you are done.