= Solution
Put $A=\mathbb C[X,Y]$, $P=(X)$ and $Q=(X^2,Y)$. The <quotient ring> $A/P\cong\mathbb C[Y]$ is an <integral domain>, so $P$ is a <prime ideal> and hence a <primary ideal>. Put $\mathfrak m=(X,Y)$. Then
$$
\mathfrak m^2=(X^2,XY,Y^2)\subseteq Q\subseteq\mathfrak m.
$$
The <polynomial ring> $A$ is a <Noetherian ring> by the <Hilbert basis theorem>, and $\mathfrak m$ is a <maximal ideal> since $A/\mathfrak m\cong\mathbb C$. Therefore the <maximal-primary ideal criterion> from <1/c> makes $Q$ $\mathfrak m$-<primary>. In particular its <radical of an ideal> is $\mathfrak m$, distinct from $P$.
Neither <primary ideal> can be omitted: $X\in P\setminus Q$, while $Y\in Q\setminus P$. Thus the given intersection is already a <minimal primary decomposition>. To express the <ideal> by generators, write an element of $P\cap Q$ as $Xh=X^2u+Yv$. Reducing modulo the <prime ideal> $(X)$ gives $Y\bar v=0$ in the <integral domain> $\mathbb C[Y]$, so $v=Xw$. Cancelling $X$ in the <integral domain> $A$ gives $h=Xu+Yw$. Hence
$$
\boxed{J=(X^2,XY)=(X)\cap(X^2,Y).}
$$
By the <first uniqueness theorem for primary decomposition>, the <associated primes of a module> $A/J$ are exactly the two <radicals of ideals>. Their strict inclusion $P\subsetneq\mathfrak m$ gives
$$
\boxed{\operatorname{Ass}_A(A/J)=\{(X),(X,Y)\},\quad (X)\text{ isolated},\quad (X,Y)\text{ embedded}.}
$$
One can also see both <associated primes of a module> directly through <annihilators>:
$$
\operatorname{Ann}_A(Y+J)=(J:Y)=(X),\qquad
\operatorname{Ann}_A(X+J)=(J:X)=(X,Y).
$$
For the first <colon ideal>, reducing $hY\in J$ modulo $(X)$ forces $h\in(X)$, and the reverse inclusion follows from $XY\in J$. For the second <colon ideal>, cancel $X$ in $hX\in X(X,Y)$. Both represented elements of the <quotient module> $A/J$ are nonzero.
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