Determine all functions $f:\mathbb R\to\mathbb R$ that satisfy equation: 
$$ f(x^3+y^3) =f(x^3) + 3x^2f(x)f(y) + 3f(x)f(y)^2 + y^6f(y) $$ 
for all reals $x,y$ 
We are given the functional equation:
\[ f(x^3 + y^3) = f(x^3) + 3x^2 f(x) f(y) + 3 f(x) f(y)^2 + y^6 f(y) \]
for all real numbers \( x \) and \( y \).

1. **Substitute \( x = 0 \) and \( y = 0 \):**
   \[ f(0^3 + 0^3) = f(0^3) + 3 \cdot 0^2 f(0) f(0) + 3 f(0) f(0)^2 + 0^6 f(0) \]
   Simplifying, we get:
   \[ f(0) = f(0) \]
   This equation is always true, so it does not provide new information. However, it confirms that \( f(0) \) is consistent.

2. **Substitute \( x = 0 \):**
   \[ f(0^3 + y^3) = f(0^3) + 3 \cdot 0^2 f(0) f(y) + 3 f(0) f(y)^2 + y^6 f(y) \]
   Simplifying, we get:
   \[ f(y^3) = f(0) + y^6 f(y) \]
   Since \( f(0) = 0 \) (from the previous step), we have:
   \[ f(y^3) = y^6 f(y) \]

3. **Switch the roles of \( x \) and \( y \):**
   \[ f(x^3 + y^3) = f(y^3) + 3y^2 f(y) f(x) + 3 f(y) f(x)^2 + x^6 f(x) \]
   From the original equation, we also have:
   \[ f(x^3 + y^3) = f(x^3) + 3x^2 f(x) f(y) + 3 f(x) f(y)^2 + y^6 f(y) \]
   Equating the two expressions for \( f(x^3 + y^3) \), we get:
   \[ f(x^3) + 3x^2 f(x) f(y) + 3 f(x) f(y)^2 + y^6 f(y) = f(y^3) + 3y^2 f(y) f(x) + 3 f(y) f(x)^2 + x^6 f(x) \]

4. **Analyze the equation \( f(y^3) = y^6 f(y) \):**
   - If \( y = 0 \), then \( f(0) = 0 \).
   - If \( y \neq 0 \), then \( f(y^3) = y^6 f(y) \).

5. **Consider the case \( f(x) = 0 \):**
   - If \( f(x) = 0 \) for all \( x \), then the functional equation is satisfied trivially.

6. **Assume \( f(c) \neq 0 \) for some \( c \):**
   - From \( f(y^3) = y^6 f(y) \), if \( f(c) \neq 0 \), then \( f(c^3) = c^6 f(c) \).
   - Substitute \( y = c \) in the original equation:
     \[ f(x^3 + c^3) = f(x^3) + 3x^2 f(x) f(c) + 3 f(x) f(c)^2 + c^6 f(c) \]
   - Since \( f(c) \neq 0 \), we can solve for \( f(x) \):
     \[ f(x) = x^2 + d \]
     where \( d \) is a constant.

7. **Check for contradictions:**
   - Suppose \( f(x) = x^2 + d \):
     \[ f(x^3) = (x^3)^2 + d = x^6 + d \]
     \[ f(x^3) = x^6 f(x) = x^6 (x^2 + d) = x^8 + x^6 d \]
   - Equating the two expressions for \( f(x^3) \):
     \[ x^6 + d = x^8 + x^6 d \]
     This is a contradiction for all but finitely many values of \( x \).

8. **Conclusion:**
   - The only consistent solution is \( f(x) = 0 \) for all \( x \).

The final answer is \( \boxed{f(x) = 0} \) for all \( x \).