Find all functions $f : [0,\infty) \to [0,\infty)$ such that $f(f(x)) +f(x) = 12x$, for all $x \ge 0$.

1. **Verification of the proposed solution:**
   We start by verifying that the function \( f(x) = 3x \) satisfies the given functional equation. 
   \[
   f(f(x)) + f(x) = 12x
   \]
   Substituting \( f(x) = 3x \):
   \[
   f(f(x)) = f(3x) = 3(3x) = 9x
   \]
   \[
   f(x) = 3x
   \]
   Adding these, we get:
   \[
   f(f(x)) + f(x) = 9x + 3x = 12x
   \]
   Thus, \( f(x) = 3x \) satisfies the equation.

2. **General solution using recurrence relation:**
   Define \( a_n = f^n(x) \) for \( n \geq 0 \), where \( f^0(x) = x \). The given functional equation can be rewritten in terms of \( a_n \):
   \[
   a_{n+2} + a_{n+1} = 12a_n
   \]
   This is a linear homogeneous recurrence relation with characteristic equation:
   \[
   r^2 + r - 12 = 0
   \]
   Solving the characteristic equation:
   \[
   r = \frac{-1 \pm \sqrt{1 + 48}}{2} = \frac{-1 \pm 7}{2}
   \]
   \[
   r_1 = 3, \quad r_2 = -4
   \]
   Therefore, the general solution to the recurrence relation is:
   \[
   a_n = c_1 \cdot 3^n + c_2 \cdot (-4)^n
   \]

3. **Non-negativity constraint:**
   Since \( f(x) \geq 0 \) for all \( x \geq 0 \), we must have \( a_n \geq 0 \) for all \( n \). The term \( c_2 \cdot (-4)^n \) will oscillate and can become negative for large \( n \) unless \( c_2 = 0 \). Thus, we must have:
   \[
   a_n = c_1 \cdot 3^n
   \]

4. **Determining the function \( f \):**
   Since \( a_n = c_1 \cdot 3^n \), we have:
   \[
   f(a_n) = a_{n+1} = c_1 \cdot 3^{n+1}
   \]
   This implies:
   \[
   f(c_1 \cdot 3^n) = c_1 \cdot 3^{n+1}
   \]
   Let \( x = c_1 \cdot 3^n \). Then:
   \[
   f(x) = 3x
   \]
   Since \( c_1 \) is an arbitrary positive real number and \( f \) is surjective, we conclude that:
   \[
   f(x) = 3x
   \]

The final answer is \( \boxed{ f(x) = 3x } \).