6. Given $x, y \neq 0$, and $x, y$ satisfy $x y=\frac{x}{y}=$ $x-y$. Then the value of $x+y$ is ( .).
(A) $-\frac{3}{2}$
(B) $-\frac{1}{2}$
(C) $\frac{1}{2}$
(D) $\frac{3}{2}$
6. A.

Since $x, y \neq 0$, from $xy = \frac{x}{y}$ we know $y^2 = 1$.
Thus, $y = \pm 1$.
If $y = 1$, from $xy = x - y$, we get $x = x - 1$, which is impossible. Therefore, $y = -1$.
From $xy = x - y$, we get $x = -\frac{1}{2}$.
Therefore, $x + y = -\frac{1}{2} + (-1) = -\frac{3}{2}$.