Find all function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for any three real number   $a,b,c$  , if $ a + f(b) + f(f(c)) = 0$ :
$$ f(a)^3 + bf(b)^2 + c^2f(c) = 3abc $$.
1. **Substitute \( a = -f(b) - f(f(c)) \) into the given equation:**

   Given the functional equation:
   \[
   f(a)^3 + bf(b)^2 + c^2f(c) = 3abc
   \]
   Substitute \( a = -f(b) - f(f(c)) \):
   \[
   f(-f(b) - f(f(c)))^3 + bf(b)^2 + c^2f(c) = 3(-f(b) - f(f(c)))bc
   \]
   Denote this assertion as \( P(b, c) \).

2. **Case I: \( f(n) + f(0) = 0 \) for all \( n \neq 0 \):**

   From \( P(0, 0) \):
   \[
   f(-f(0) - f(f(0)))^3 = 0 \implies f(-f(0) - f(f(0))) = 0
   \]
   We will prove that \( f(0) = 0 \) by contradiction.

   Suppose \( f(0) \neq 0 \). Then, \( f(f(0)) + f(0) = 0 \). This implies:
   \[
   f(-f(0) - f(f(0))) = 0 \implies f(0) = 0
   \]
   which is a contradiction. Therefore, we must have \( f(0) = 0 \). Hence, \( f(n) + f(0) = 0 \implies f(n) = 0 \) for all \( n \neq 0 \). This gives us \( f(x) = 0 \) as one possible solution to the functional equation.

3. **Case II: There exists \( n \neq 0 \) such that \( f(n) + f(0) \neq 0 \):**

   From \( P(0, 0) \):
   \[
   f(-f(0) - f(f(0)))^3 = 0 \implies f(-f(0) - f(f(0))) = 0
   \]
   From \( P(-f(0) - f(f(0)), 0) \):
   \[
   f(-f(f(0)))^3 = 0 \implies f(-f(f(0))) = 0
   \]
   Now, we know that there exists \( n \neq 0 \) such that \( f(n) + f(0) \neq 0 \). This means, \( (f(n) + f(0))n \neq 0 \).

   Let \( k \) and \( \ell \) be any real numbers such that \( f(k) = f(\ell) = 0 \). From \( P(n, k) \) and \( P(n, \ell) \):
   \[
   f(-f(n) - f(0))^3 + nf(n)^2 + 3(f(n) + f(0))nk = 0
   \]
   \[
   f(-f(n) - f(0))^3 + nf(n)^2 + 3(f(n) + f(0))n\ell = 0
   \]
   This shows us that:
   \[
   3(f(n) + f(0))nk = 3(f(n) + f(0))n\ell \implies k = \ell
   \]
   since \( (f(n) + f(0))n \neq 0 \).

   Since \( f(-f(0) - f(f(0))) = 0 \) and \( f(-f(f(0))) = 0 \), this means \( -f(0) - f(f(0)) = -f(f(0)) \implies f(0) = 0 \).

   Furthermore, if \( k \) is a real number such that \( f(k) = 0 \), then this also means that \( k = 0 \).

   From \( P(0, c) \):
   \[
   f(-f(f(c)))^3 + c^2f(c) = 0
   \]
   Also, from \( P(f(c), 0) \):
   \[
   f(-f(f(c)))^3 + f(c)f(f(c))^2 = 0
   \]
   This shows us that \( f(f(c))^2 = c^2 \) for all real numbers \( c \).

4. **Claim: \( f(f(c)) = c \) for all real numbers \( c \):**

   We know that for all real numbers \( c \), \( f(f(c))^2 = c^2 \). This means, either \( f(f(c)) = c \) or \( f(f(c)) = -c \).

   We already know that if \( c = 0 \), then \( f(f(0)) = 0 \). Now, we prove it for \( c \neq 0 \).

   Suppose there exists \( c \neq 0 \) such that \( f(f(c)) = -c \). Then from \( P(0, c) \):
   \[
   f(c)^3 + cf(c)^2 = 0 \implies f(c)^2 + c^2 = 0
   \]
   This implies that \( c = 0 \), which is a contradiction, because we assumed that \( c \neq 0 \).

   Therefore, \( f(f(c)) = c \) for all \( c \neq 0 \).

5. **From \( P(f(b), c) \) and the fact that \( f(f(c)) = c \) for all real numbers \( c \):**

   \[
   f(-b - c)^3 + b^2f(b) + c^2f(c) = 3(-b - c)f(b)c
   \]
   and from \( P(b, f(c)) \):
   \[
   f(-b - c)^3 + b^2f(b) + c^2f(c) = 3(-b - c)bf(c)
   \]
   Therefore, \( f(b)c = bf(c) \) for all \( b + c \neq 0 \). We have:
   \[
   \frac{f(b)}{b} = \frac{f(c)}{c} \implies \frac{f(b)}{b} = k
   \]
   for some constant \( k \). Therefore, \( f(b) = bk \) for all real numbers \( b \), where \( k \) is some constant.

6. **Finally, from \( f(f(b)) = b \implies bk^2 = b \), we have \( k = 1 \) or \( k = -1 \):**

   This gives us two more solutions, \( f(x) = x \) and \( f(x) = -x \).

The final answer is \( \boxed{f(x) = x} \), \( \boxed{f(x) = -x} \), and \( \boxed{f(x) = 0} \).