2. Transform each equation by bringing the fractions to a common denominator. We obtain

$$
\left\{\begin{array}{l}
(\sqrt{2}+1)(x-y)=\sqrt{2}\left(x^{2}+y^{2}\right)-3 x y \\
(\sqrt{2}-1)(x-y)=\sqrt{2}\left(x^{2}+y^{2}\right)+3 x y
\end{array}\right.
$$

if $x \neq \pm \sqrt{2} y$ and $y \neq \pm \sqrt{2} x$. Adding and subtracting these equations, we find

$$
\left\{\begin{array}{c}
x-y=x^{2}+y^{2} \\
y-x=3 x y
\end{array}\right.
$$

Therefore, $x^{2}+3 x y+y^{2}=0$, from which either $x=y=0$ (this option does not fit due to the conditions stated above), or $y=\frac{-3 \pm \sqrt{5}}{2} x \neq 0$. Then from the second equation, we find $\frac{-3 \pm \sqrt{5}}{2} \cdot 3 x^{2}=\frac{-5 \pm \sqrt{5}}{2} x$, so $x=$ $\frac{-5 \pm \sqrt{5}}{3 \cdot(-3 \pm \sqrt{5})}=\frac{5 \pm \sqrt{5}}{6}$, and then $y=\frac{-5 \pm \sqrt{5}}{6}$ respectively (here the signs are taken either both upper or both lower).
Answer: $(x, y)=((5+\sqrt{5}) / 6,(-5+\sqrt{5}) / 6) ;((5-\sqrt{5}) / 6,(-5-\sqrt{5}) / 6)$. Answer to the variant: $5-2:(x, y)=((3+\sqrt{3}) / 8,(-3+\sqrt{3}) / 8) ;(x, y)=$ $(3-\sqrt{3}) / 8,(-3-\sqrt{3}) / 8)$.