<s>[INST] $\forall x\in\mathbb R$, $|x|\neq 1$ it is known that $f\left(\frac{x-3}{x+1}\right)+f\left(\frac{3+x}{1-x}\right)=x$. Find $f(x)$. 
$\forall x\in\mathbb R$, $|x|\neq 1$ $$f\left(\frac{x-3}{x+1}\right)+f\left(\frac{3+x}{1-x}\right)=x$$Find $f(x)$.

Now what I'm actually looking for is an explanation of a solution to this problem. I haven't really ever had any experience with such equations.

The solution:
Let $t=\frac{x-3}{x+1}$. Then $$f(t)+f\left(\frac{t-3}{t+1}\right)=\frac{3+t}{1-t}$$
Now let $t=\frac{3+x}{1-x}$. Then $$f\left(\frac{3+t}{1-t}\right)+f(t)=\frac{t-3}{t+1}$$
Add both equalities: $$\frac{8t}{1-t^2}=2f(t)+f\left(\frac{t-3}{t+1}\right)+f\left(\frac{3+t}{1-t}\right)=2f(t)+t$$
Hence the answer is $$f(x)=\frac{4x}{1-x^2}-\frac{x}{2}$$

This is unclear to me. For instance, how come we can assign a different value to the same variable? Does anyone understand this? I'd appreciate any help.
 [/INST] It's a bit unclear, but if you look for what $t$'s each of the equalities hold, it starts to make sense.
Note that if $x\ne-1$ and $t_0\ne1$, then $t_0=\frac{x-3}{x+1}\Longleftrightarrow(x+1)t_0=x-3\Longleftrightarrow t_0+3=x(1-t_0)\Longleftrightarrow x=\frac{3+t_0}{1-t_0}$
So the first equality holds for all $t_0\in\mathbb R\setminus\{1\}$.
Similarly if $x\ne1$ and $t_1\ne-1$, then $t_1=\frac{3+x}{1-x}\Longleftrightarrow(1-x)t_1=3+x\Longleftrightarrow t_1-3=x(1+t_1)\Longleftrightarrow x=\frac{t_1-3}{t_1+1}$
So the second equality holds for all $t_1\in\mathbb R\setminus\{-1\}$.
Now for any $x\in\mathbb R$, $|x|\ne1$ we know that $$f(x)+f\left(\frac{x-3}{x+1}\right)=\frac{3+x}{1-x}$$ by the first equation, substituting $x$ for $t_0$ and $$f\left(\frac{3+x}{1-x}\right)+f(x)=\frac{x-3}{x+1}$$ by the second equation, substituting $x$ for $t_1$.
Now add these to get the solution.
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