6. Given positive real numbers $x, y$ satisfy
$$
\frac{1}{x+3 y}+\frac{1}{2 x+y}=1 \text {. }
$$

Then the minimum value of $x+y$ is
6. $\frac{3+2 \sqrt{2}}{5}$.
$$
\begin{array}{l}
\text { Given } 1=\frac{1}{x+3 y}+\frac{1}{2 x+y}=\frac{1}{x+3 y}+\frac{2}{4 x+2 y} \\
\geqslant \frac{(1+\sqrt{2})^{2}}{5(x+y)} \\
\Rightarrow x+y \geqslant \frac{(1+\sqrt{2})^{2}}{5}=\frac{3+2 \sqrt{2}}{5} .
\end{array}
$$

When $x=\frac{4+\sqrt{2}}{10}, y=\frac{2+3 \sqrt{2}}{10}$, the equality holds.
Thus, the minimum value of $x+y$ is $\frac{3+2 \sqrt{2}}{5}$.