6. The integer solutions $(x, y)$ of the equation $\frac{1}{y}+\frac{1}{2(x+y)}=\frac{1}{y^{2}}$ are ( ).
(A) 0 sets
(B) 1 set
(C) finite sets (more than 1 set)
(D) infinitely many sets
6. B
The equation is transformed into: $(x+2 y)^{2}=(x+y)[(x+y)+2]$.
(1) When the left side is 0, one of the factors on the right side must be 0. Solving yields $x=-4, y=2$ and $x=0, y=0$ (the extraneous solution is discarded).
(2) When the left side is not 0, it is a perfect square. However, on the right side, when $x+y>0$, $(x+y)^{2}<(x+y)^{2}+2(x+y)<(x+y+1)^{2}$ is not a perfect square;
when $x+y<0$,
$$
\begin{aligned}
(|x+y|-2)^{2} & <(x+y)^{2}-2|x+y| \\
& <(|x+y|-1)^{2}
\end{aligned}
$$

is also not a perfect square.
Therefore, the equation has only one solution $(-4,2)$.