Find all functions $ f: \mathbb{R} \mapsto \mathbb{R}$ such that $ \forall x,y,z \in \mathbb{R}$ we have: If
\[ x^3 \plus{} f(y) \cdot x \plus{} f(z) \equal{} 0,\]
then
\[ f(x)^3 \plus{} y \cdot f(x) \plus{} z \equal{} 0.\]
1. **Claim 1:** There exists a real number \( x_0 \) such that \( f(x_0) = 0 \).

   **Proof:** Consider the equation \( x^3 + x f(0) + f(0) = 0 \). This is a cubic polynomial in \( x \), and by the Fundamental Theorem of Algebra, it has at least one real root. Let \( x_0 \) be such a root. Then, \( x_0^3 + x_0 f(0) + f(0) = 0 \). Since \( (x_0, 0, 0) \) is a valid triple, we have \( f(x_0)^3 + 0 \cdot f(x_0) + 0 = 0 \), which simplifies to \( f(x_0)^3 = 0 \). Therefore, \( f(x_0) = 0 \). \(\blacksquare\)

2. **Claim 2:** \( f(0) = 0 \).

   **Proof:** Consider the valid triple \( (0, y, x_0) \) where \( x_0 \) satisfies \( f(x_0) = 0 \). Then, we have \( f(0)^3 + y f(0) + x_0 = 0 \). Since \( f(x_0) = 0 \), this simplifies to \( f(0)^3 + y f(0) = 0 \). For this to hold for all \( y \in \mathbb{R} \), it must be that \( f(0) = 0 \). \(\blacksquare\)

3. **Claim 3:** \( f \) is injective.

   **Proof:** Suppose \( f(a) = f(b) \). Consider a valid triple \( (x, y, a) \). Then, \( x^3 + f(y) x + f(a) = 0 \). Since \( f(a) = f(b) \), \( (x, y, b) \) is also valid, so \( x^3 + f(y) x + f(b) = 0 \). This implies \( f(x)^3 + y f(x) + a = 0 \) and \( f(x)^3 + y f(x) + b = 0 \). Therefore, \( a = b \), proving injectivity. \(\blacksquare\)

4. **Claim 4:** \( f \) is surjective.

   **Proof:** Fix an arbitrary \( t \in \mathbb{R} \). Consider the polynomial \( P(X) = (X - t)(X^2 + tX + (1 + t^2)) \). Set \( z_0 = -t(1 + t^2) \) and \( y_0 = 1 \). Let \( x_0 \) be a real number such that \( (x_0, y_0, z_0) \) is valid. Then, \( P(x_0) = f(x_0)^3 + y_0 f(x_0) + z_0 = 0 \). Since \( P \) has exactly one real root \( t \), we conclude \( f(x_0) = t \), establishing surjectivity. \(\blacksquare\)

5. **Claim 5:** For all \( y \ge 0 \), we have \( f(y) \ge 0 \).

   **Proof:** Suppose for contradiction that there exists \( y_0 > 0 \) such that \( f(y_0) < 0 \). Let \( x_0 \) be a solution to \( x^2 = -f(y_0) \). Then, \( (x_0, y_0, 0) \) is valid, so \( f(x_0)^3 + y_0 f(x_0) = 0 \). This cannot occur since \( x_0 \neq 0 \) implies \( f(x_0) \neq 0 \) by injectivity, and \( f(x_0)^2 + y_0 > 0 \). This contradiction implies \( f(y) \ge 0 \) for all \( y \ge 0 \). \(\blacksquare\)

6. **Claim 6:** \( f \) is additive.

   **Proof:** Fix any three distinct real numbers \( a, b, c \) such that \( a + b + c = 0 \). Since \( f \) is surjective, we can choose real numbers \( y_0, z_0 \) such that \( a, b, c \) are the roots of the polynomial \( f(X) = X^3 + X f(y) + f(z) = 0 \). Since \( f(a), f(b), f(c) \) are distinct real numbers, they are the roots of the equation \( X^3 + y X + z = 0 \). It follows that \( f(a) + f(b) + f(c) = 0 \). For all real numbers \( a, b \), we get \( f(-(a + b)) = -(f(a) + f(b)) \). Setting \( b = 0 \), we get \( f(-x) = -f(x) \) for all \( x \), yielding \( f(a + b) = f(a) + f(b) \) for all distinct reals \( a, b \). For \( a = b \), consider \( f(2a) = f(a + m) + f(a - m) = (f(a) + f(m)) + (f(a) - f(m)) = 2f(a) \), where \( m \) is sufficiently large. Thus, \( f \) is completely additive. \(\blacksquare\)

7. **Claim 7:** \( f \) is the identity function.

   **Proof:** By Claims 5 and 6, \( f \) is additive and bounded on an interval, so \( f \) is linear. Since \( f(0) = 0 \), we get \( f(x) = cx \) for all \( x \). Notice that \( c \neq 0 \), so from a valid triple \( (1, y, z) \), we get \( z = \left(\frac{-1}{c} - y\right) = -c^3 - cy \). This fails to hold for \( y \rightarrow \infty \) unless \( c = 1 \), in which case we are done. \(\blacksquare\)

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