If $f$ and $g$ are differentiable at $x_{0}$, and $g$ is non-zero on $X$, then $f/g$ is also differentiable at $x_{0}$ and \begin{align*} \left(\frac{f}{g}\right)' = \frac{f'(x_{0})g(x_{0}) - f(x_{0})g'(x_{0})}{g(x_{0})^{2}} \end{align*}
MY ATTEMPT
According to the definition of derivative, we have that \begin{align*} \lim_{x\rightarrow x_{0}}\frac{(f/g)(x) - (f/g)(x_{0})}{x-x_{0}} = \lim_{x\rightarrow x_{0}}\frac{f(x)g(x_{0}) - f(x_{0})g(x)}{(x-x_{0})g(x)g(x_{0})} \end{align*} On the other hand, we have that \begin{align*} f(x)g(x_{0}) - f(x_{0})g(x) & = f(x)g(x_{0}) - f(x_{0})g(x_{0}) + f(x_{0})g(x_{0}) - f(x_{0})g(x)\\\\ & = [f(x) - f(x_{0})]g(x_{0}) - f(x_{0})[g(x) - g(x_{0})] \end{align*}
Consequently, we have that \begin{align*} \lim_{x\rightarrow x_{0}}\frac{(f/g)(x) - (f/g)(x_{0})}{x-x_{0}} & = \lim_{x\rightarrow x_{0}}\frac{[f(x) - f(x_{0})]g(x_{0}) - f(x_{0})[g(x) - g(x_{0})] }{(x-x_{0})g(x)g(x_{0})}\\\\ & = \lim_{x\rightarrow x_{0}}\frac{[f(x) - f(x_{0})]g(x_{0})}{(x-x_{0})g(x)g(x_{0})} - \lim_{x\rightarrow x_{0}}\frac{f(x_{0})[g(x) - g(x_{0})]}{(x-x_{0})g(x)g(x_{0})}\\\\ & = \frac{f'(x_{0})g(x_{0}) - f(x_{0})g'(x_{0})}{g(x_{0})^{2}} \end{align*}
and we are done.
I would like to know if someone could provide an alternative solution to this problem. Any contribution is appreciated.