If $$f(x-y)=f(x) \circ g(y)-f(y) \circ g(x)$$ And $$g(x-y) =g(x) \circ g(y)+f(x) \circ f(y)$$ for all $$x, y \in \mathrm{R}$$. If right-hand derivative at $$x=0$$ exists for $$f(x)$$, find the derivative of $$g(x)$$ at $$x=0$$