Prove the existence of limit of $x_{n+1}=x_n+\frac{x_n^2}{n^2}$
Let $x_1 = t \in (0,1)$. For any $n \ge 1$, it is easy to see $x_n \ge 0$. Since $e^y \ge 1 + y$ for all $y$, we have $$x_n = x_1 \prod_{k=1}^{n-1} \frac{x_{k+1}}{x_k} = t \prod_{k=1}^{n-1} \left(1 + \frac{x_{k}}{k^2}\right) \le t\exp\left(\sum_{k=1}^{n-1} \frac{x_k}{k^2}\right)\tag{*1}$$
Notice
- $x_1 \le t$.
- if $x_n \le tn$, then $x_{n+1} \le n t + t^2 \le (n+1)t$.
By induction, we have $x_n \le n t$ for all $n$.
Substitute this into $(*1)$, we get
$$x_n \le t \exp\left(t\sum_{k=1}^{n-1}\frac{1}{k}\right) \le t\exp\left(t(\log n + \gamma)\right) \le e^\gamma n^t$$ where $\gamma$ is Euler-Mascheroni constant.
Substitute above into $(*1)$ again, we find $$x_n \le t \exp\left(e^{\gamma} \sum_{k=1}^{n-1}\frac{1}{k^{2-t}}\right) \le t \exp\left( e^\gamma \zeta(2-t)\right) < \infty$$ where $\zeta(t)$ is the Riemann Zeta function.
Because the sequence is increasing $$x_{n+1}=x_{n}+\frac{x_{n}^2}{n^2}<x_{n}+\frac{x_{n}x_{n+1}}{n^2}$$ $$\frac{1}{x_{n}}<\frac{1}{x_{n+1}}+\frac{1}{n^2}$$ (where it is easy to show that the $x_{n}>0$ is positive for all $n$)
so, $$\sum_{i=1}^n(\frac{1}{x_{i}}-\frac{1}{x_{i+1}})<\sum_{i=1}^n\frac{1}{i^2}$$ Because of $$\sum_{i=1}^n\frac{1}{i^2}<1+\sum_{i=2}^n\frac{1}{(i-1)i}<2-\frac{1}{n}<2$$ $$\sum_{i=1}^n(\frac{1}{x_{i}}-\frac{1}{x_{i+1}})<2$$ $$\frac{1}{x_{1}}-\frac{1}{x_{n+1}}<2$$ which is equivalent to $$x_{n+1}<\frac{1}{\frac{1}{x_{1}}-2}$$ so the sequence is bounded above
Since $0\lt x_1\lt1$, the relation $$ x_{n+1}=x_n+\frac{x_n^2}{n^2}\tag{1} $$ implies not only that $x_n$ is increasing, but also inductively that $$ 0\lt x_n\lt n\tag{2} $$ Equation $(1)$ implies $$ \begin{align} \left(\frac1{x_{n+1}}-\frac1{n+1}\right)-\left(\frac1{x_n}-\frac1n\right) &=\left(\frac1n-\frac1{n+1}\right)-\left(\frac{n^2+x_n}{n^2x_n+x_n^2}-\frac{n^2}{n^2x_n+x_n^2}\right)\\ &=\frac1{n^2+n}-\frac1{n^2+x_n}\\ &=-\frac{n-x_n}{(n^2+n)(n^2+x_n)}\\ &\ge-\frac{n-x_n}{n^3(n+1)}\\ &\ge-\frac{n-x_n}{x_nn^2(n+1)}\\ &=-\frac1{n(n+1)}\left(\frac1{x_n}-\frac1n\right)\tag{3} \end{align} $$ Therefore, $$ \left(\frac1{x_{n+1}}-\frac1{n+1}\right) \ge\left(1-\frac1{n(n+1)}\right)\left(\frac1{x_n}-\frac1n\right)\tag{4} $$ and since $$ \prod_{n=1}^\infty\left(1-\frac1{n(n+1)}\right)=\frac1\pi\sin\left(\frac\pi\phi\right)\tag{5} $$ we have $$ \frac1{x_n}-\frac1n\ge\frac1\pi\sin\left(\frac\pi\phi\right)\left(\frac1{x_1}-1\right)\tag{6} $$ which means $$ \bbox[5px,border:2px solid #C0A000]{x_n\le\pi\csc\left(\frac\pi\phi\right)\frac{x_1}{1-x_1}}\tag{7} $$ where $\pi\csc\left(\frac\pi\phi\right)=3.3706903036$.
Therefore, since $x_n$ is increasing and bounded above, the limit exists and is bounded by $(7)$.
Motivational Note
I looked at $$ x_{n+1}-x_n=\frac{x_n^2}{n^2}\tag{8} $$ as representative of $$ \frac{\mathrm{d}x}{\mathrm{d}n}=\frac{x^2}{n^2}\tag{9} $$ whose solution is $$ \frac1x-\frac1n=C\tag{10} $$ So I considered $\frac1{x_n}-\frac1n$ which lead to $(3)$, $(4)$, $(6)$ and $(7)$.
Derivation of $\boldsymbol{(5)}$
$$
\begin{align}
\prod_{k=1}^n\left(1-\frac1{k(k+1)}\right)
&=\prod_{k=1}^n\frac{(k+1-\phi)(k+\phi)}{k(k+1)}\tag{11}\\
&=\underbrace{\frac{\Gamma(n+2-\phi)}{\Gamma(2-\phi)}}_{\prod(k+1-\phi)}
\underbrace{\frac{\Gamma(n+1+\phi)}{\Gamma(1+\phi)}}_{\prod(k+\phi)}
\underbrace{\frac{\Gamma(1)}{\Gamma(n+1)}}_{\prod\frac1k}
\underbrace{\frac{\Gamma(2)}{\Gamma(n+2)}}_{\prod\frac1{k+1}}\tag{12}\\
&\to\frac1{\Gamma(2-\phi)\Gamma(1+\phi)}\tag{13}\\[6pt]
&=\frac1{\Gamma\left(1-\frac1\phi\right)\Gamma\left(2+\frac1\phi\right)}\tag{14}\\
&=\frac1{\Gamma\left(1-\frac1\phi\right)\Gamma\left(\frac1\phi\right)}\tag{15}\\
&=\frac1\pi\sin\left(\frac\pi\phi\right)\tag{16}
\end{align}
$$
Explanation:
$(11)$: $k^2+k-1=(k+1-\phi)(k+\phi)$
$(12)$: break up the product of factors into a product of Gamma functions
$(13)$: $\lim\limits_{n\to\infty}\frac{\Gamma(n+2-\phi)\Gamma(n+1+\phi)}{\Gamma(n+1)\Gamma(n+2)}=1$ by Gautschi's Inequality
$(14)$: $\phi=1+\frac1\phi$
$(15)$: $\Gamma\left(2+\frac1\phi\right)=\left(1+\frac1\phi\right)\frac1\phi\,\Gamma\left(\frac1\phi\right)=\phi\,\frac1\phi\,\Gamma\left(\frac1\phi\right)=\Gamma\left(\frac1\phi\right)$
$(16)$: Euler's Reflection Formula