Find all functions $\displaystyle f : \mathbb{R} - \left\{ -\frac{1}{3}, \frac{1}{3} \right\} \rightarrow \mathbb{R}$ that satisfy the identity
$\displaystyle f \left(\frac{x+1}{1-3x} \right) = x-f(x)$
Consider the change of variables: $\displaystyle
x : \to {\textstyle{{x - 1} \over {1 + 3x}}}
$
Then we get: $\displaystyle
f\left( x \right) = {\textstyle{{x - 1} \over {1 + 3x}}} - f\left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right)
$ ( check that the denominator is never 0)
So we've got these 2 equations: $\displaystyle
\left[ \begin{array}{l}
f\left( x \right) = {\textstyle{{x - 1} \over {1 + 3x}}} - f\left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right) \\
f\left( {{\textstyle{{x + 1} \over {1 - 3x}}}} \right) = x - f\left( x \right) \\
\end{array} \right.
$
Thus: $\displaystyle
f\left( {{\textstyle{{x + 1} \over {1 - 3x}}}} \right) - f\left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right) = x - {\textstyle{{x - 1} \over {1 + 3x}}} = {\textstyle{{3x^2 + 1} \over {3x + 1}}}
$
Again take: $\displaystyle
x: \to {\textstyle{{x - 1} \over {1 + 3x}}}
$
Then: $\displaystyle
f\left( x \right) - f\left( {{\textstyle{{x + 1} \over {1 - 3x}}}} \right) = {\textstyle{{3 \cdot \left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right)^2 + 1} \over {3 \cdot \left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right) + 1}}}
$
Consider then the system of equations: $\displaystyle
\left\{ \begin{array}{l}
f\left( x \right) - f\left( {{\textstyle{{x + 1} \over {1 - 3x}}}} \right) = {\textstyle{{3 \cdot \left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right)^2 + 1} \over {3 \cdot \left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right) + 1}}} \\
f\left( {{\textstyle{{x + 1} \over {1 - 3x}}}} \right) = x - f\left( x \right) \\
\end{array} \right.
$
and solve for $\displaystyle
f\left( x \right)
$