2. The eccentricity of the conic section $2 x^{2}+3 x y+2 y^{2}=1$ is $\qquad$ .
2. $\frac{\sqrt{42}}{7}$.
$$
\begin{array}{l}
\text { Given } 2 x^{2}+3 x y+2 y^{2}=1 \\
\Rightarrow \frac{1}{2}\left(x^{2}+y^{2}\right)+\frac{3}{2}(x+y)^{2}=1 \\
\Rightarrow x^{2}+y^{2} \leqslant 2 .
\end{array}
$$

It can be seen that the quadratic curve is an ellipse centered at the origin.
Furthermore, when $x+y=0$ and $x^{2}+y^{2}=2$, the distance from the point $(x, y)$ to the origin is maximized, which is $\sqrt{2}$, i.e., the semi-major axis $a=\sqrt{2}$.
$$
\begin{array}{l}
\text { Given } 2 x^{2}+3 x y+2 y^{2}=1 \\
\Rightarrow \frac{7}{2}\left(x^{2}+y^{2}\right)-\frac{3}{2}(x-y)^{2}=1 \\
\Rightarrow x^{2}+y^{2} \geqslant \frac{2}{7} .
\end{array}
$$

When $x=y$ and $x^{2}+y^{2}=\frac{2}{7}$, the distance from the point $(x, y)$ to the origin is minimized, which is $\sqrt{\frac{2}{7}}$, i.e., the semi-minor axis $b=\sqrt{\frac{2}{7}}$. Therefore, $e=\sqrt{\frac{a^{2}-b^{2}}{a^{2}}}=\frac{\sqrt{42}}{7}$.