If $f(\frac{x+y}{x-y})=\frac{f(x)+f(y)}{f(x)-f(y)}$, which of the following statement is correct

Let $f : \mathbb{Z} \to \mathbb{R}$ solve the functional equation

$$ \forall x, y \in \mathbb{Z} \ \ \text{s.t.} \ \ \frac{x+y}{x-y} \in \mathbb{Z}, \quad f\left(\frac{x+y}{x-y}\right) = \frac{f(x) + f(y)}{f(x) - f(y)}. \tag{*} $$

Step 1. $f(1) = 1$ and $f(0) = 0$.

For $x \neq 0$, plug $y=0$ to see that $f(1)=\frac{f(x)+f(0)}{f(x)-f(0)}$ holds. If $f(1) \neq 1$, then solving in terms of $f(x)$ gives

$$ f(x) = f(0) \frac{f(1) + 1}{f(1) - 1}, $$

and so, $f$ is constant on $\mathbb{Z}\setminus\{0\}$. This immediately yields a contradiction to $\text{(*)}$, since plugging $x = 2$ and $y = 1$ renders the right-hand side of $\text{(*)}$ undefined. This forces $f(1) = 1$, which then implies $f(0) = 0$.

Step 2. $f(-x) = -f(x)$.

In view of Step 1, it suffices to assume $x \neq 0$. Plugging $y = -x$, we have

$$ 0 = f(0) = \frac{f(x) + f(-x)}{f(x) - f(-x)}, $$

and so, $f(x) + f(-x) = 0$ and the desired claim follows.

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By Step 1 and 2, (1), (2) and (3) are correct. Also, notice that $f(x) = x$ solves $\text{(*)}$ but does not satisfy (4). So (4) cannot be true.