Let a sequence $(x_n)$ satisfy :$x_1=1$ and $x_{n+1}=x_n+3\sqrt{x_n} + \frac{n}{\sqrt{x_n}}$,$\forall$n$\ge1$
a) Prove lim$\frac{n}{x_n}=0$
b) Find lim$\frac{n^2}{x_n}$
### Part (a): Prove \(\lim_{n \to \infty} \frac{n}{x_n} = 0\)

1. **Initial Setup and Induction Hypothesis**:
   - Given \(x_1 = 1\) and \(x_{n+1} = x_n + 3\sqrt{x_n} + \frac{n}{\sqrt{x_n}}\).
   - We need to show that \(\lim_{n \to \infty} \frac{n}{x_n} = 0\).

2. **Induction Base Case**:
   - For \(n = 1\), \(x_1 = 1\).
   - \(\frac{1}{x_1} = 1\).

3. **Induction Step**:
   - Assume \(\frac{n}{x_n} \to 0\) as \(n \to \infty\).
   - We need to show \(\frac{n+1}{x_{n+1}} \to 0\).

4. **Analyzing the Recurrence Relation**:
   - From the recurrence relation, \(x_{n+1} = x_n + 3\sqrt{x_n} + \frac{n}{\sqrt{x_n}}\).
   - Since \(x_n\) is increasing, \(x_{n+1} > x_n\).

5. **Bounding \(x_n\)**:
   - We have \(x_{n+1} \ge x_n + 3\sqrt{x_n}\).
   - By induction, \(x_n \ge n^2\).

6. **Limit Calculation**:
   - \(\frac{n}{x_n} \le \frac{n}{n^2} = \frac{1}{n}\).
   - As \(n \to \infty\), \(\frac{1}{n} \to 0\).

Thus, \(\lim_{n \to \infty} \frac{n}{x_n} = 0\).

### Part (b): Find \(\lim_{n \to \infty} \frac{n^2}{x_n}\)

1. **Claim 1: \(x_n \ge n^2\)**:
   - **Base Case**: For \(n = 1\), \(x_1 = 1 \ge 1^2\).
   - **Induction Step**: Assume \(x_n \ge n^2\).
     \[
     x_{n+1} = x_n + 3\sqrt{x_n} + \frac{n}{\sqrt{x_n}} \ge n^2 + 3n \ge (n+1)^2.
     \]
   - Therefore, \(x_n \ge n^2\) for all \(n\).

2. **Claim 2: \(x_n \le \frac{9n^2}{4}\)**:
   - **Base Case**: For \(n = 1\), \(x_1 = 1 \le \frac{9 \cdot 1^2}{4}\).
   - **Induction Step**: Assume \(x_n \le \frac{9n^2}{4}\).
     \[
     x_{n+1} = x_n + 3\sqrt{x_n} + \frac{n}{\sqrt{x_n}} \le \frac{9n^2}{4} + \frac{9n}{2} + 1 \le \frac{9}{4}(n+1)^2.
     \]
   - Therefore, \(x_n \le \frac{9n^2}{4}\) for all \(n\).

3. **Limit Calculation**:
   - From Claim 1 and Claim 2, we have:
     \[
     n^2 \le x_n \le \frac{9n^2}{4}.
     \]
   - Dividing by \(n^2\):
     \[
     1 \le \frac{x_n}{n^2} \le \frac{9}{4}.
     \]
   - Taking the limit as \(n \to \infty\):
     \[
     \lim_{n \to \infty} \frac{x_n}{n^2} = \frac{9}{4}.
     \]

Thus, \(\lim_{n \to \infty} \frac{n^2}{x_n} = \frac{4}{9}\).

The final answer is \( \boxed{ \frac{4}{9} } \).