Here's my attempt. As first suggested by JG89, first show that \frac{2 - (x_n - 1)^2}{3} < 1 holds. This will show that x_n is bounded above by 1 for all n \in \mathbb{N}.
Next, we need to show that x_n \geq 0 for all n \in \mathbb{N}. For contradiction, suppose there exists some N+1 \in \mathbb{N} such that x_{N+1} < 0. So we have that 0 < \frac{1 + 2x_N - x_N^2}{3}, implying that 0 < 2 - (x_N - 1)^2. Do some further rearranging, get 2 > (x_N - 1)^2 \geq 0 and finally, get \sqrt{2} + 1 > x_N \geq 1 --- contradiction to the prior where we had shown that x_n < 1 for all n \in \mathbb{N}.
Next, we show that the sequence (x_n) is monotonically increasing. To show that x_{n+1} \geq x_n for all n is equivalent to showing x_{n+1} - x_n \geq 0. So, consider that x_{n+1} - x_n = \frac{1 + 2x_n - x_n^2}{3} - x_n = \frac{1 + 2x_n - x_n^2 - 3x_n}{3} = \frac{1 - x_n - x_n^2}{3} \geq \frac{1 - x_n}{3} > 0 and the last inequality holds because we had previously shown that 0 \leq x_n < 1 holds.
Hence, by the monotone convergence theorem, the sequence (x_n) converges.
In fact, we can do better than that --- we will find the limit. Note that if \lim x_n = x, then \lim x_{n+1} = x (this is easy to prove). Hence, from the original recurrence relation, we can have x^2 = \frac{1 + 2x - x^2}{3}. Using the typical quadratic formula, we solve to get x = \frac{-1 \pm \sqrt{5}}{2} and we reject the negative root since we had shown that x_n \geq 0, which implies \lim x_n \geq 0. Hence, \lim x_n = (-1 + \sqrt{5})/2.
