Let $A=k[x,y]$ where $k$ is an algebraically closed field and let $M=A/(xy)$ be an $A$-module. I am supposed to calculate $\text{Supp}(M)= \{ P \in \text{Spec}(A) : M_p \not= 0 \}$ where $M_p = S^{-1}M, S=A \setminus P$. I am not sure about this, but I am thinking if $M_p \not= 0$ then it exist an $a/b \not= 0/1$, that is, it doesnt exist a $u \in k[x,y] \setminus P$ with $ua=0$ but this is only true if $( (x) \cup (y) \cup (xy)) \subset P$, is this correct?
Conversely, if $( (x) \cup (y) \cup (xy)) \subset P$ then of course $M_p ≠ 0$, to see this, take any $a/b \in M_p$ with $a ≠ 0 + (xy)$ then $a/b ≠ 0/1$, because if $a/b=0/1$ then it must exist a $u \in k[x,y] \setminus P$ with $ua=0$ but then $ua$ must contain the product $xy$ but this is only true if $a$ contains $xy$ which is not the case since $a ≠ 0+ (xy)$.
In summary $\text{Supp}(M)= \{ P \in \text{Spec}(A) : ( (x) \cup (y) \cup (xy)) \subset P \}$. I feel I am missing something, is this correct?
It is almost correct, except for taking into account that $(xy) \subset (x)$, and likewise $(xy) \subset (y)$. Also $P$ contains $(xy)$ if and only if $P$ contains $(x)$ or $P$ contains $(y)$ since $P$ is a prime ideal, so that $(x) \cup (y) \cup (xy)=(x) \cup (y)$, and $$\DeclareMathOperator\supp{Supp} \DeclareMathOperator\spec{Spec}\supp(M)=\Bigl\{ P \in \spec(A)\: :\: P \supset (x)\enspace \text{or}\enspace P\supset (y)\Bigr\}.$$