1. Given that $x$ and $y$ are integers, and satisfy
$$
\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}\right)=-\frac{2}{3}\left(\frac{1}{x^{4}}-\frac{1}{y^{4}}\right) \text {. }
$$

Then the number of possible values for $x+y$ is $(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 4
$-1 . C$.
From the given equation, we have
$$
\frac{x+y}{x y} \cdot \frac{x^{2}+y^{2}}{x^{2} y^{2}}=\frac{2}{3} \cdot \frac{x^{4}-y^{4}}{x^{4} y^{4}}
$$

Obviously, $x$ and $y$ are not equal to 0.
Thus, $x+y=0$ or $3 x y=2(x-y)$.
If $3 x y=2(x-y)$, then
$$
(3 x+2)(3 y-2)=-4 \text{. }
$$

Since $x, y$ are integers, solving yields $(x, y)=(-1,2)$ or $(-2,1)$.
Therefore, $x+y=1$ or -1.
In summary, there are 3 possible values for $x+y$.