a. We will prove $x_n \ge n^2$ for all positive integers by induction. The statement is obvious in case $n=1$. Assume that $x_n \ge n^2$, hence
$$x_{n+1} > x_n + 3\sqrt{x_n} \ge x_n + 3n \ge (n+1)^2.$$
Therefore, the statement is proved. Thus
$$0 < \frac{n}{x_n} \geq \frac{1}{n}.$$
By the Squeeze theorem, one can get $\lim_{n \to +\infty} \frac{n}{x_n} = 0$.

b. Let $x_n = y_n^2$ and the given formula can be rewritten as
$$(y_{n+1} - y_n)(y_{n+1} + y_n) = 3y_n + \frac{n}{y_n}$$
then combine with the formula of $y_{n+1}$ calculated by $y_n$, the above
equality obtains that
$$\begin{align*} y_{n+1} - y_n &= \frac{3y_n + \frac{n}{y_n}}{\sqrt{y_n^2 + 3y_n + \frac{n}{y_n}} + y_n} \\ &= \frac{3 + \frac{n}{y_n^2}}{\sqrt{1 + \frac{3}{y_n} + \frac{n}{y_n^3}} + 1} \end{align*}$$
From part a), one could deduce that
$$\lim_{n \to +\infty} \frac{1}{y_n} = \lim_{n \to +\infty} \frac{n}{y_n^3} = 0$$
hence
$$\lim_{n \to +\infty} (y_{n+1} - y_n) = \frac{3}{2}.$$
By Cesàro mean theorem,
$$\lim_{n \to +\infty} \frac{y_n}{n} = \frac{3}{2} \Rightarrow \lim_{n \to +\infty} \frac{n^2}{x_n} = \frac{4}{9}$$
and it is the result.