Find all $f: \mathbb{Q}_{+} \rightarrow \mathbb{R}$ such that \[ f(x)+f(y)+f(z)=1 \] holds for every positive rationals $x, y, z$ satisfying $x+y+z+1=4xyz$.
1. **Transform the given equation:**

   The given equation is:
   \[
   x + y + z + 1 = 4xyz
   \]
   We can rewrite this equation in terms of new variables. Let:
   \[
   u = \frac{1}{2x+1}, \quad v = \frac{1}{2y+1}, \quad w = \frac{1}{2z+1}
   \]
   Then, we need to show that:
   \[
   \frac{1}{2x+1} + \frac{1}{2y+1} + \frac{1}{2z+1} = 1
   \]
   This transformation is valid because:
   \[
   \frac{1}{2x+1} + \frac{1}{2y+1} + \frac{1}{2z+1} = 1 \implies u + v + w = 1
   \]

2. **Define a new function:**

   Define a bijective function \( g: \mathbb{Q}^{+} \to \mathbb{Q} \cap (0,1) \) by:
   \[
   g(x) = \frac{1}{2x+1}
   \]
   Our condition now becomes: For \( x, y, z \in \mathbb{Q}^{+} \) with \( g(x) + g(y) + g(z) = 1 \), we have \( f(x) + f(y) + f(z) = 1 \).

3. **Introduce a new function \( h \):**

   Let:
   \[
   h(x) = f(g^{-1}(x)) - \frac{1}{3}
   \]
   Then the condition is equivalent to: For \( x, y, z \in (0, 1) \) with \( x + y + z = 1 \), we have:
   \[
   h(x) + h(y) + h(z) = 0
   \]

4. **Simplify the problem:**

   From the condition \( h(x) + h(y) + h(z) = 0 \), we can derive:
   \[
   h(0) = 0
   \]
   and:
   \[
   h(-x) = -h(x)
   \]
   for \( 0 < x < 1 \).

5. **Solve the Cauchy equation:**

   The problem is now equivalent to solving the Cauchy equation for \( h \) on the interval \( (0, 1) \):
   \[
   h(x + y) = h(x) + h(y)
   \]
   for \( 0 \leq x, y < 1 \).

6. **Determine the form of \( h \):**

   Let \( h(1/2) = a/2 \) for some \( a \in \mathbb{R} \). Starting from:
   \[
   h\left(\frac{1}{2n}\right) + h\left(\frac{1}{2n}\right) = h\left(\frac{2}{2n}\right)
   \]
   we get:
   \[
   h\left(\frac{m}{2n}\right) = m \times h\left(\frac{1}{2n}\right)
   \]
   for natural numbers \( m \leq n \). Since \( h(1) = a \), we have:
   \[
   h\left(\frac{m}{2n}\right) = a \times \frac{m}{2n}
   \]
   for \( m \leq n \).

7. **Generalize \( h \):**

   For any \( 0 < x = \frac{p}{q} < 1 \), we have:
   \[
   h\left(\frac{p}{q}\right) = a \times \frac{p}{q}
   \]
   From \( h(-x) = -h(x) \) and \( h(0) = 0 \), we get:
   \[
   h(x) = ax
   \]
   for \( -1 < x < 1 \).

8. **Return to the original function \( f \):**

   Returning back to \( f \), we get:
   \[
   f(x) = \frac{1}{3} + a \left( \frac{1}{2x+1} - \frac{1}{3} \right) = a \times \frac{1}{2x+1} + (1-a) \frac{1}{3}
   \]

The final answer is \( \boxed{ f(x) = a \times \frac{1}{2x+1} + (1-a) \frac{1}{3} } \)