6. Given real numbers $x, y$ satisfy $x^{2}-x y-2 y^{2}=1$. Then the minimum value of $2 x^{2}+y^{2}$ is $\qquad$ .


6. $\frac{2+2 \sqrt{3}}{3}$.

Notice,
$$
\begin{array}{l}
x^{2}-x y-2 y^{2}=1 \Leftrightarrow(x+y)(x-2 y)=1 . \\
\text { Let } x+y=m, x-2 y=n .
\end{array}
$$

Then $x=\frac{2 m+n}{3}, y=\frac{m-n}{3}$.
Thus $2 x^{2}+y^{2}=\frac{1}{9}\left(9 m^{2}+3 n^{2}+6 m n\right)$
$$
=\frac{1}{9}\left(9 m^{2}+\frac{3}{m^{2}}+6\right) \geqslant \frac{2+2 \sqrt{3}}{3} \text {. }
$$

Equality holds if and only if $m= \pm \frac{1}{\sqrt[4]{3}}, n= \pm \sqrt[4]{3}$.