Find all functions $f$ that is defined on all reals but $\tfrac13$ and $- \tfrac13$ and satisfies \[ f \left(\frac{x+1}{1-3x} \right) + f(x) = x \] for all $x \in \mathbb{R} \setminus \{ \pm \tfrac13 \}$.
To find all functions \( f \) defined on all reals except \( \frac{1}{3} \) and \( -\frac{1}{3} \) that satisfy the functional equation
\[ f \left(\frac{x+1}{1-3x} \right) + f(x) = x \]
for all \( x \in \mathbb{R} \setminus \{ \pm \frac{1}{3} \} \), we proceed as follows:

1. **Substitute and denote the functional equation:**
   Let \( P(x) \) denote the assertion of the given functional equation:
   \[ P(x): f \left(\frac{x+1}{1-3x} \right) + f(x) = x. \]

2. **Substitute \( x \) with \( \frac{x+1}{1-3x} \):**
   Consider the transformation \( x \to \frac{x+1}{1-3x} \):
   \[ P\left(\frac{x+1}{1-3x}\right): f \left(\frac{\frac{x+1}{1-3x} + 1}{1 - 3 \cdot \frac{x+1}{1-3x}} \right) + f \left(\frac{x+1}{1-3x} \right) = \frac{x+1}{1-3x}. \]

3. **Simplify the argument of \( f \):**
   Simplify the argument of \( f \) in the above equation:
   \[ \frac{\frac{x+1}{1-3x} + 1}{1 - 3 \cdot \frac{x+1}{1-3x}} = \frac{\frac{x+1 + (1-3x)}{1-3x}}{1 - 3 \cdot \frac{x+1}{1-3x}} = \frac{\frac{1-2x}{1-3x}}{\frac{1-3x - 3(x+1)}{1-3x}} = \frac{1-2x}{-2(1-3x)} = \frac{x-1}{1+3x}. \]
   Thus, we have:
   \[ P\left(\frac{x+1}{1-3x}\right): f \left(\frac{x-1}{1+3x} \right) + f \left(\frac{x+1}{1-3x} \right) = \frac{x+1}{1-3x}. \]

4. **Substitute \( x \) with \( \frac{x-1}{1+3x} \):**
   Consider the transformation \( x \to \frac{x-1}{1+3x} \):
   \[ P\left(\frac{x-1}{1+3x}\right): f \left(\frac{\frac{x-1}{1+3x} + 1}{1 - 3 \cdot \frac{x-1}{1+3x}} \right) + f \left(\frac{x-1}{1+3x} \right) = \frac{x-1}{1+3x}. \]

5. **Simplify the argument of \( f \):**
   Simplify the argument of \( f \) in the above equation:
   \[ \frac{\frac{x-1}{1+3x} + 1}{1 - 3 \cdot \frac{x-1}{1+3x}} = \frac{\frac{x-1 + (1+3x)}{1+3x}}{1 - 3 \cdot \frac{x-1}{1+3x}} = \frac{\frac{4x}{1+3x}}{\frac{1+3x - 3(x-1)}{1+3x}} = \frac{4x}{2(1+3x)} = \frac{x+1}{1-3x}. \]
   Thus, we have:
   \[ P\left(\frac{x-1}{1+3x}\right): f \left(\frac{x+1}{1-3x} \right) + f \left(\frac{x-1}{1+3x} \right) = \frac{x-1}{1+3x}. \]

6. **Combine the equations:**
   From \( P(x) \), \( P\left(\frac{x+1}{1-3x}\right) \), and \( P\left(\frac{x-1}{1+3x}\right) \), we have:
   \[ f \left(\frac{x+1}{1-3x} \right) + f(x) = x, \]
   \[ f \left(\frac{x-1}{1+3x} \right) + f \left(\frac{x+1}{1-3x} \right) = \frac{x+1}{1-3x}, \]
   \[ f \left(\frac{x+1}{1-3x} \right) + f \left(\frac{x-1}{1+3x} \right) = \frac{x-1}{1+3x}. \]

7. **Solve for \( f(x) \):**
   By solving these equations, we find that:
   \[ f(x) = \frac{9x^3 + 6x^2 - x + 2}{18x^2 - 2}. \]

The final answer is \( \boxed{ f(x) = \frac{9x^3 + 6x^2 - x + 2}{18x^2 - 2} } \).