My Solution
Notice that $$x_n+\dfrac{4}{x_{n+1}^2}<3=3\sqrt[3]{\frac{x_n}{2}\cdot\frac{x_n}{2}\cdot\frac{4}{x_n^2}}\leq \frac{x_n}{2}+\frac{x_n}{2}+\frac{4}{x_n^2}=x_n+\frac{4}{x_n^2}.$$ This shows that $$x_{n+1}>x_n.\tag1$$ Moreover $$x_n<3-\frac{4}{x_{n+1}^2}<3.\tag2$$ From $(1)$ and $(2)$, according to the monotone convergence theorem, $x_n$ is convergent,namely,$\lim\limits_{n \to \infty}x_n$ exists, whom we denote as $L$.
Since $$x_n+\dfrac{4}{x_{n+1}^2}<3,$$ taking the limits,we obtain $$L+\frac{4}{L^2}\leq 3. \tag3$$ Similarily, since $$x_n+\dfrac{4}{x_{n}^2}\geq 3,$$ taking the limits,we obtain $$L+\frac{4}{L^2}\geq 3. \tag4$$ From $(3)$ and $(4)$, we obtain $$L+\frac{4}{L^2}=3.$$ Thus, we can solve to have $$L=2.$$ The other root $L=-1$ contradicts $x_n>0,$ which implies $L \geq 0$.
Please correct me if I'm wrong. Hope to see other solutions.