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.