why this function is even function 
Let $f(x)$ be a continuous differentiable odd function, and for any
  $x,y\in R$, such  $$f(x-y)=f(x)g(y)-g(x)f(y)$$
show that $g(x)$ is an even function.

How can I go about solving this? I think there is a specific function
$$\sin{(x-y)}=\sin{x}\cos{y}-\cos{x}\sin{y}$$
Thanks

This isn't true, even if you assume $f$ is not identically zero.  For instance, when $f(x)=\sin x$, the function $g(x)=\sin x+\cos x$ satisfies the equation but is not even.  More generally, if $g(x)$ is any function that satisfies that equation, so is $g(x)+cf(x)$ for any constant $c$.
However, assuming that $f'(0)\neq 0$ and $g'(0)=0$, you can prove $g$ is even as follows.  First, show that $g(0)=1$ (this is easy).  Next, solve the equation for $g(x)$, and then take the limit as $y\to 0$ using L'Hopital's rule.  This gives you $g(x)=f'(x)/f'(0)$.  Since $f$ is odd, $f'$ is even, so $g$ is even.  More generally, if you don't assume $g'(0)=0$, you can conclude $g(x)=cf(x)+f'(x)/f'(0)$ for some constant $c=g'(0)/f'(0)$.
