Find all functions $f:R -\{0,2\} \to R$ that satisfy for all $x \ne 0,2$ $$f(x) \cdot \left(f\left(\sqrt[3]{\frac{2+x}{2-x}}\right) \right)^2=\frac{x^3}{4}$$
1. **Restate the problem correctly:**
   We need to find all functions \( f: \mathbb{R} \setminus \{0, 2\} \to \mathbb{R} \) that satisfy the functional equation:
   \[
   f(x) \cdot \left(f\left(\sqrt[3]{\frac{2+x}{2-x}}\right)\right)^2 = \frac{x^3}{4}
   \]
   for all \( x \neq 0, 2 \).

2. **Substitute \( x = 0 \):**
   \[
   f(0) \cdot \left(f\left(\sqrt[3]{\frac{2+0}{2-0}}\right)\right)^2 = \frac{0^3}{4} = 0
   \]
   This simplifies to:
   \[
   f(0) \cdot \left(f(1)\right)^2 = 0
   \]
   Since the product is zero, either \( f(0) = 0 \) or \( f(1) = 0 \).

3. **Case 1: \( f(0) = 0 \):**
   - Substitute \( x = -2 \):
     \[
     f(-2) \cdot \left(f\left(\sqrt[3]{\frac{2+(-2)}{2-(-2)}}\right)\right)^2 = \frac{(-2)^3}{4} = -2
     \]
     This simplifies to:
     \[
     f(-2) \cdot \left(f(0)\right)^2 = -2
     \]
     Since \( f(0) = 0 \), we get:
     \[
     f(-2) \cdot 0 = -2 \implies 0 = -2
     \]
     This is a contradiction. Therefore, \( f(0) \neq 0 \).

4. **Case 2: \( f(1) = 0 \):**
   - Substitute \( x = 1 \):
     \[
     f(1) \cdot \left(f\left(\sqrt[3]{\frac{2+1}{2-1}}\right)\right)^2 = \frac{1^3}{4} = \frac{1}{4}
     \]
     This simplifies to:
     \[
     0 \cdot \left(f\left(\sqrt[3]{3}\right)\right)^2 = \frac{1}{4}
     \]
     Since \( f(1) = 0 \), we get:
     \[
     0 = \frac{1}{4}
     \]
     This is also a contradiction. Therefore, \( f(1) \neq 0 \).

Since both cases lead to contradictions, there is no function \( f \) that satisfies the given functional equation.

The final answer is \(\boxed{\text{No such function}}\)