If the equation:

$f(\frac{x-3}{x+1}) + f(\frac{3+x}{1-x}) = x$

holds true for all real x but $\pm 1$, find $f(x)$.
1. Define the transformations \( u(x) = \frac{x-3}{x+1} \) and \( v(x) = \frac{3+x}{1-x} \). We need to verify the properties of these transformations:
   \[
   u(v(x)) = u\left(\frac{3+x}{1-x}\right) = \frac{\left(\frac{3+x}{1-x}\right) - 3}{\left(\frac{3+x}{1-x}\right) + 1} = \frac{\frac{3+x-3(1-x)}{1-x}}{\frac{3+x+1-x}{1-x}} = \frac{\frac{3+x-3+3x}{1-x}}{\frac{4}{1-x}} = \frac{4x}{4} = x
   \]
   Similarly,
   \[
   v(u(x)) = v\left(\frac{x-3}{x+1}\right) = \frac{3 + \frac{x-3}{x+1}}{1 - \frac{x-3}{x+1}} = \frac{\frac{3(x+1) + (x-3)}{x+1}}{\frac{(x+1) - (x-3)}{x+1}} = \frac{\frac{3x + 3 + x - 3}{x+1}}{\frac{x + 1 - x + 3}{x+1}} = \frac{4x}{4} = x
   \]
   Thus, \( u(v(x)) = v(u(x)) = x \).

2. Next, we need to find \( f(x) \) such that:
   \[
   f\left(\frac{x-3}{x+1}\right) + f\left(\frac{3+x}{1-x}\right) = x
   \]
   Let \( u(x) = \frac{x-3}{x+1} \) and \( v(x) = \frac{3+x}{1-x} \). Then:
   \[
   f(u(x)) + f(v(x)) = x
   \]

3. Consider the equation \( f(u(u(x))) + f(x) = u(x) \):
   \[
   f\left(u\left(\frac{x-3}{x+1}\right)\right) + f(x) = \frac{x-3}{x+1}
   \]
   Since \( u(u(x)) = v(x) \), we have:
   \[
   f(v(x)) + f(x) = u(x)
   \]

4. Similarly, consider the equation \( f(x) + f(v(v(x))) = v(x) \):
   \[
   f(x) + f\left(v\left(\frac{3+x}{1-x}\right)\right) = \frac{3+x}{1-x}
   \]
   Since \( v(v(x)) = u(x) \), we have:
   \[
   f(x) + f(u(x)) = v(x)
   \]

5. Adding the two equations:
   \[
   f(u(x)) + f(v(x)) + f(x) + f(x) = u(x) + v(x)
   \]
   Simplifying, we get:
   \[
   f(u(x)) + f(v(x)) + 2f(x) = u(x) + v(x)
   \]
   Using \( f(u(x)) + f(v(x)) = x \), we substitute:
   \[
   x + 2f(x) = u(x) + v(x)
   \]
   Solving for \( f(x) \):
   \[
   2f(x) = u(x) + v(x) - x
   \]
   \[
   f(x) = \frac{u(x) + v(x) - x}{2}
   \]

6. Calculate \( u(x) + v(x) \):
   \[
   u(x) = \frac{x-3}{x+1}, \quad v(x) = \frac{3+x}{1-x}
   \]
   \[
   u(x) + v(x) = \frac{x-3}{x+1} + \frac{3+x}{1-x}
   \]
   Finding a common denominator:
   \[
   u(x) + v(x) = \frac{(x-3)(1-x) + (3+x)(x+1)}{(x+1)(1-x)} = \frac{x - 3 - x^2 + 3x + 3x + x^2 + 3}{x^2 - 1} = \frac{4x}{x^2 - 1}
   \]
   Thus:
   \[
   f(x) = \frac{\frac{4x}{x^2 - 1} - x}{2} = \frac{4x - x(x^2 - 1)}{2(x^2 - 1)} = \frac{4x - x^3 + x}{2(x^2 - 1)} = \frac{-x^3 + 5x}{2(x^2 - 1)}
   \]
   Simplifying further:
   \[
   f(x) = -\frac{x^3 - 5x}{2(x^2 - 1)} = -\frac{x(x^2 - 1 + 4)}{2(x^2 - 1)} = -\frac{x(x^2 + 3)}{2(x^2 - 1)}
   \]

The final answer is \( \boxed{ f(x) = -\frac{x^3 + 7}{2(x^2 - 1)} } \).