1. Given that $x, y$ are integers, and satisfy $\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}\right)=-\frac{2}{3}\left(\frac{1}{x^{4}}-\frac{1}{y^{4}}\right)$. Then the possible values of $x+y$ are ( ) .
(A) 1
(B) 2
(C) 3
(D) 4
-1. .
From the given equation, we have
$$
\frac{x+y}{x y} \cdot \frac{x^{2}+y^{2}}{x^{2} y^{2}}=\frac{2}{3} \cdot \frac{x^{4}-y^{4}}{x^{4} y^{4}} \text {. }
$$

Clearly, $x$ 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 $x$ and $y$ are integers, we solve to get
$$
(x, y)=(-1,2) \text { or }(-2,1) \text {. }
$$

Therefore, $x+y=1$ or -1.
In summary, there are 3 possible values for $x+y$.