You can easily show that $(x_n)$ is bounded below by 0 and bounded above by 1.
You can then show (by induction e.g.) that $(x_{2n})$ is decreasing and that $(x_{2n+1})$ is increasing. Then you can argue that the sequence $(x_{2n})$ converges to some number $L$ and that the sequence $(x_{2n+1})$ converges to some number $M$.
Now, since $x_{n+1} ={1-x_{n+1} x_n\over3}$, it follows that $(x_n)$ converges, to $b$, say. Then from the recursion formula, we must have $b={1\over b+3}$; solving this equation we see that $b$ is its positive solution $b={-3\over2}+{\sqrt{13}\over2}$.
For the induction argument to show that $(x_{2n})$ is decreasing and $(x_{2n+1})$ is increasing:
Verify that $x_1 and that $x_2>x_4$.
Assume that both $x_{2n-1} and $x_{2n-2}>x_{2n}$ hold.
Then show that $x_{2n+2}. Using this result, show that $x_{2n+3}>x_{2n+1}$.