We are given the function $f(x) = \frac{x-1}{x+1}$. We need to prove that $f(2x) = \frac{3f(x)+1}{f(x)+3}$.

1. Problem Description

We are given the function $f(x) = \frac{x-1}{x+1}$. We need to prove that $f(2x) = \frac{3f(x)+1}{f(x)+3}$.

2. Solution Steps

First, let's find $f(2x)$:

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

Now, let's find $\frac{3f(x)+1}{f(x)+3}$:

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

$3f(x) + 1 = 3\left(\frac{x-1}{x+1}\right) + 1 = \frac{3(x-1) + (x+1)}{x+1} = \frac{3x - 3 + x + 1}{x+1} = \frac{4x - 2}{x+1}$

$f(x) + 3 = \frac{x-1}{x+1} + 3 = \frac{x-1 + 3(x+1)}{x+1} = \frac{x - 1 + 3x + 3}{x+1} = \frac{4x + 2}{x+1}$

$\frac{3f(x)+1}{f(x)+3} = \frac{\frac{4x-2}{x+1}}{\frac{4x+2}{x+1}} = \frac{4x-2}{4x+2} = \frac{2(2x-1)}{2(2x+1)} = \frac{2x-1}{2x+1}$

Since $f(2x) = \frac{2x - 1}{2x + 1}$ and $\frac{3f(x)+1}{f(x)+3} = \frac{2x-1}{2x+1}$, we have proven that $f(2x) = \frac{3f(x)+1}{f(x)+3}$.

3. Final Answer

$f(2x) = \frac{3f(x)+1}{f(x)+3}$ is proven.