Let $x_{0}, x_{1}, x_{2}, \ldots$ be a sequence of positive real numbers such that for all $n \geq 0$,
$$
x_{n+1}=\frac{\left(n^{2}+1\right) x_{n}^{2}}{x_{n}^{3}+n^{2}} \text {. }
$$

For which values of $x_{0}$ is this sequence bounded?
First consider the derivative of the function $f_{n}(x)=\frac{\left(n^{2}+1\right) x_{n}^{2}}{x_{n}^{3}+n^{2}}$.
$$
\begin{aligned}
f_{n}^{\prime}(x) & =\frac{2\left(n^{2}+1\right) x\left(x^{3}+n^{2}\right)-\left(n^{2}+1\right) x^{2}\left(3 x^{2}\right)}{\left(x^{3}+n^{2}\right)^{2}} \\
& =\frac{\left(n^{2}+1\right) x}{\left(x^{3}+n^{2}\right)^{2}} \times\left(2 n^{2}-x^{3}\right) .
\end{aligned}
$$

So there is a unique critical point at $x=\sqrt[3]{2 n^{2}}$ and so $f_{n}(x)$ is increasing on the interval $\left(0, \sqrt[3]{2 n^{2}}\right)$ and decreasing on the interval $\left(\sqrt[3]{2 n^{2}}, \infty\right)$.
We now let $x_{0}=\lambda$ and find the following first few terms.
$$
\begin{array}{l}
x_{0}=\lambda \\
x_{1}=\frac{\left(0^{2}+1\right) \lambda^{2}}{\lambda^{3}+0^{2}}=\lambda^{-1} \\
x_{2}=\frac{\left(1^{2}+1\right) x_{1}^{2}}{x_{1}^{3}+1^{2}}=\frac{2 \lambda^{-2}}{\lambda^{-3}+1}=\frac{2 \lambda}{1+\lambda^{3}}
\end{array}
$$

Observation One: For any $n \geq 2$, if $x_{n} \leq 1$ then $x_{n+1} \leq 1$.
Proof: Since $f_{n}(x)$ is increasing on the interval $01$. Note that $x_{2}=\frac{2 \lambda}{1+\lambda^{3}}\frac{n^{2}}{2}$ and $\left(n^{2}+b^{3}\right)\frac{x_{n}\left(x_{n}-1\right) \frac{n^{2}}{2}}{2 n^{2}}=\frac{x_{n}\left(x_{n}-1\right)}{4}>0,
$$
meaning the sequence $\left(x_{n}\right)$ is increasing for $n \geq N$. Letting $a=x_{N}>1$, we can conclude that for all $n \geq N$,
$$
x_{n+1}-x_{n} \geq \frac{a(a-1)}{4} .
$$

Therefore the sequence is eventually always greater than an arithmetic progression with common difference $d=\frac{a(a-1)}{4}$, which contradicts the assumption that $b$ was an upper bound.
So the sequence is unbounded if and only if $x_{2}>1$.
Solving the inequality $x_{2}=\frac{2 x_{0}}{x_{0}^{3}+1}>1$ yields the final answer that the sequence is bounded if and only if $0<x_{0} \leq \frac{\sqrt{5}-1}{2}$ or $x_{0} \geq 1$.