34.3. Find all functions $f(x)$ that are defined for all $x \neq 0, \pm 1$ and satisfy the relation

$$
x f(x) + 2 f\left(\frac{x-1}{x+1}\right) = 1
$$

## 34.2. Functional Equations for Arbitrary Functions
34.3. For $\varphi_{1}(x)=\frac{x-1}{x+1}$, we sequentially find: $\varphi_{2}(x)=-1 / x$, $\varphi_{3}(x)=\frac{x+1}{1-x}$, and $\varphi_{4}(x)=x$. Therefore, we obtain the system of equations

$$
\left\{\begin{array}{l}
x f(x)+2 f\left(\frac{x-1}{x+1}\right)=1 \\
\frac{x-1}{x+1} f\left(\frac{x-1}{x+1}\right)+2 f\left(-\frac{1}{x}\right)=1 \\
-\frac{1}{x} f\left(-\frac{1}{x}\right)+2 f\left(\frac{x+1}{1-x}\right)=1 \\
\frac{x+1}{1-x} f\left(\frac{x+1}{1-x}\right)+2 f(x)=1
\end{array}\right.
$$

To reduce the calculations, we proceed as follows. Multiply the third equation by $2 x$ and add it to the second equation. Then add to the resulting equation the first equation, multiplied by $\frac{1-x}{2(x+1)}$, and the last equation, multiplied by $\frac{4 x(x-1)}{x+1}$. As a result, we get

$$
\frac{15 x(x-1)}{2(x+1)} f(x)=\frac{12 x^{2}-3 x+3}{2(x+1)}
$$

Thus, if $x \neq 0, \pm 1$, then $f(x)=\frac{4 x^{2}-x+1}{5(x-1)}$. Direct verification shows that this function $f(x)$ satisfies the required relation.