An injective module $M$ over a domain $D$ is divisible: this means that for every non-zero $d\in D$ and every $m\in M$ there exists $n\in N$ with $m=d\cdot n$.
But here, choosing $m=1\in K=M$, it is impossible to write $1=x\cdot n\in K$ since for every $n\in K$ one actually has $x\cdot n=0$
So $K$ is not injective.
1) The $K[x,y]$-module $K$ is of finite presentation. Hence it is flat if and only if it is projective.
Since I already proved that $K$ is not projective, it is not flat either.
2) Here is another proof, in the vein of Martin's comment.
If $I\subset R$ is an ideal in a ring, for the quotient $R/I$
to be flat over $R$ it is necessary that $I=I^2$.
Since $(x,y)\neq(x,y)^2$, $K=K[x,y]/(x,y)$ is not flat over $K[x,y]$.
3) A third proof is obtained by remembering that a flat module over a domain is torsion-free.
Since $x\cdot 1=0\in K$ even though $x\neq 0\in K[x,y]$ and $1\neq 0\in K$, the module $K$ has torsion and is thus not flat.
As a generalization of the above, it might be worth remembering that if $0\subsetneq I\subsetneq R$ is a non-trivial ideal of a domain, the $R$-module $R/I$ is never projective, nor injective, nor flat.