$\mathbb{Q}$ is set of all rational numbers. Find all functions $f:\mathbb{Q}\times\mathbb{Q}\rightarrow\mathbb{Q}$ such that for all $x$, $y$, $z$ $\in\mathbb{Q}$  satisfy 
$f(x,y)+f(y,z)+f(z,x)=f(0,x+y+z)$
1. Let $\mathbb{Q}$ be the set of all rational numbers. We need to find all functions $f: \mathbb{Q} \times \mathbb{Q} \rightarrow \mathbb{Q}$ such that for all $x, y, z \in \mathbb{Q}$, the following equation holds:
   \[
   f(x,y) + f(y,z) + f(z,x) = f(0, x+y+z)
   \]

2. Define the assertion $P(x, y, z)$ as:
   \[
   P(x, y, z): f(x, y) + f(y, z) + f(z, x) = f(0, x+y+z)
   \]

3. Let $g(x) = f(0, x)$. Then, $P(0, 0, 0)$ implies:
   \[
   f(0, 0) + f(0, 0) + f(0, 0) = f(0, 0) \implies 3f(0, 0) = f(0, 0) \implies f(0, 0) = 0
   \]
   Hence, $g(0) = 0$.

4. Consider $P(x, 0, 0)$:
   \[
   f(x, 0) + f(0, 0) + f(0, x) = f(0, x) \implies f(x, 0) = 0
   \]

5. Consider $P(x, y, 0)$:
   \[
   f(x, y) + f(y, 0) + f(0, x) = f(0, x+y) \implies f(x, y) + 0 + g(x) = g(x+y) \implies f(x, y) = g(x+y) - g(x)
   \]

6. Substitute $f(x, y) = g(x+y) - g(x)$ into the original assertion $P(x, y, z)$:
   \[
   (g(x+y) - g(x)) + (g(y+z) - g(y)) + (g(z+x) - g(z)) = g(x+y+z)
   \]
   Simplifying, we get:
   \[
   g(x+y) + g(y+z) + g(z+x) - g(x) - g(y) - g(z) = g(x+y+z)
   \]

7. Let $Q(x, y, z)$ be the new assertion:
   \[
   Q(x, y, z): g(x+y) + g(y+z) + g(z+x) = g(x+y+z) + g(x) + g(y) + g(z)
   \]

8. Consider $Q((n+1)x, x, -x)$:
   \[
   g((n+2)x) + g(x) + g(-x) = g((n+1)x + x - x) + g((n+1)x) + g(x) + g(-x)
   \]
   Simplifying, we get:
   \[
   g((n+2)x) = 2g((n+1)x) - g(nx) + g(x) + g(-x)
   \]

9. Let $h_1(x) = g(x) + g(-x)$. The previous equation implies:
   \[
   h_1(nx) = n^2 h_1(x)
   \]
   Hence, $h_1(x) = h_1(1)x^2$ for all $x \in \mathbb{Q}$.

10. Let $h_2(x) = g(x) - g(-x)$. The previous equation implies:
   \[
   h_2(nx) = n h_2(x)
   \]
   Hence, $h_2(x) = h_2(1)x$ for all $x \in \mathbb{Q}$.

11. Therefore, we can write:
   \[
   g(x) = \frac{h_1(x) + h_2(x)}{2} = \frac{h_1(1)x^2 + h_2(1)x}{2}
   \]
   Let $a = \frac{h_1(1)}{2}$ and $b = \frac{h_2(1)}{2}$. Then:
   \[
   g(x) = ax^2 + bx
   \]

12. Substituting back into $f(x, y) = g(x+y) - g(x)$, we get:
   \[
   f(x, y) = a(x+y)^2 + b(x+y) - (ax^2 + bx) = ay^2 + 2axy + by
   \]

The final answer is $\boxed{f(x, y) = ay^2 + 2axy + by}$ for all $x, y \in \mathbb{Q}$, where $a, b \in \mathbb{Q}$.