Conjecture function $g(x)$ is even function? Let $f,g:R\to R\setminus\{0\}$ and $\forall x,y\in R$,such
$$\color{crimson}{f(x-y)=f(x)g(y)-f(y)g(x)}$$ I have  prove  the function $\color{crimson}f$ odd function.

because let $y=0$ we have
$$f(x)=f(x)g(0)-f(0)g(x)\tag{1}$$
Let $x=0,y=x$ we have
$$f(-x)=f(0)g(x)-f(x)g(0)\tag{2}$$
by $(1),(2)$ ,then $$f(x)=-f(-x)$$

Conjecture： $\color{crimson}{g(x)}$ is even function

For this function $g(x)$ is even problem ,I don't have any idea to prove it.But I think is right,because
$$\color{blue}{\sin{(x-y)}=\sin{x}\cos{y}-\sin{y}\cos{x}}$$
$$\color{crimson}{\sinh{(x-y)}=\sinh{x}\cosh{y}-\sinh{y}\cosh{x}}$$
The conjecture is false.

Consider $f(x) = x$ and $g(x) = 1+x$.
Now, for all $x,y \in \mathbb{R}$,
$$f(x-y) = x-y$$
and
$$f(x)g(y) - f(y)g(x) = x(1+y) - y(1+x) = x + xy - y - xy = x-y,$$
so $f(x-y) = f(x)g(y) - f(y)g(x)$. However, $g$ is not even.
You're in the right direction when you got $Eq-2$ by putting $x=0$ and $y=x$. From that you get $g(x) = \frac{f(-x) + f(x)g(0)}{f(0)}$. Similarly, putting $x=0$ and $y=-x$ can give you $g(-x)$ which is same as $g(x)$. Thus proving $g(x)$ is even. Note that dividing by $f(0)$ is safe because the range of $f$ does not include 0.

BTW, the function range for odd functions usually include 0, unless you don't want to define your function to be defined at 0. You could argue that $\lim_{x \rightarrow 0} f(x) = 0$ but then I am talking things that I am not sure about :D