Proof of quotient rule $(\frac{f}{g})'(x_{0})=\frac{f'(x_0)g(x_0)-f(x_0)g'(x_0)}{g^2(x_0)}$ $$\left(\frac{f}{g}\right)'(x_{0})=\frac{f'(x_0)g(x_0)-f(x_0)g'(x_0)}{g^2(x_0)}$$
So, $\frac{1}{g}.f=\frac{f}{g}$, then $$\frac{f}{g}'(x_0)=\frac{f(x)\frac{1}{g(x)}-f(x_0)\frac{1}{g(x_0)}}{x-x_0}=f(x)\frac{\frac{1}{g(x)}-\frac{1}{g(x_0)}}{x-x_0}-\frac{1}{g(x_0)}\frac{f(x)-f(x_0)}{x-x_0}$$ I took the limit of everything as $x \to x_0$ $$=f(x_0)\frac{1}{g'(x_0)}-\frac{1}{g(x_0)}f'(x_0)=\frac{f(x_0)}{g'(x_0)}-\frac{f'(x_0)}{g(x_0)}=\frac{g(x_0)f(x_0)-f'(x_0)g'(x_0)}{g'(x_0)g(x_0)}$$ which clearly it's fake.
Where did I go wrong?
Thank you!

There is another way to do it : logarithmic differentiation
$$y=\frac{f(x)}{g(x)}\implies \log(y)=\log(f(x))-\log(g(x))$$ Differentiate both sides
$$\frac{y'}y=\frac{f'(x)}{f(x)}-\frac{g'(x)}{g(x)}=\frac{f'(x)\,g(x)-f(x)\,g'(x) } {f(x)\,g(x) }$$
$$y'=y \,\frac{y'}y=\frac{f(x)}{g(x)}\times \frac{f'(x)\,g(x)-f(x)\,g'(x) } {f(x)\,g(x) }$$ Simplify.
