Does L'Hopital's rule imply that $\lim_{x\to a}\frac{f'(x)}{g'(x)} = \lim_{x\to a}\frac{f(x)}{g(x)}$ always?

The standard theorem says that both limits are equal if

1. $f$ and $g$ are differentiable on an interval containing $a$, with the possible exception of $a$ itself
2. $\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0$ or $\lim_{x \to a} |f(x)| = \infty$ and $\lim_{x \to a} |g(x)| = \infty$
3. $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ exists.

You can use it to find the limit of $f(x)/g(x)$ from the limit of $f'(x)/g'(x)$ or to find the limit of $f'(x)/g'(x)$ from the limit of $f(x)/g(x)$, but you do need to know that the limit of $f'(x)/g'(x)$ exists.

EDIT: Your "apparent proof" is bogus because it violates assumption (2).

The answer to your first question is No.

Take $$f(x)=x^2\sin(\frac 1x)$$ and $$g(x)=x$$

then

$$\lim_{x\to 0}\frac{f(x)}{g(x)}=0$$

But

$$\lim_{x\to0}\frac{f'(x)}{g'(x)}$$ Does not exist.

For clarity, let's let $F(x)=f(x)/g(x)$ and $G(x)=g(x)/f(x)$. The identity you call attention to can be written as

$${F'(x)\over G'(x)}=-{f(x)^2\over g(x)^2}$$

and we also, of course, have

$${f(x)^2\over g(x)^2}={f(x)/g(x)\over g(x)/f(x)}={F(x)\over G(x)}$$

Now if L'Hopital's rule applied to the limit for $F/G$, then we would have

$$\lim{f(x)^2\over g(x)^2}=\lim{F(x)\over G(x)}\color{red}{=}\lim{F'(x)\over G'(x)}=-\lim{f(x)^2\over g(x)^2}$$

(where the red equal sign is the L'Hopital step). But in order to apply L'Hopital to the ratio $F/G$, we must have either $\lim F(x)=\lim G(x)=0$ or $\lim F(x)=\lim G(x)=\pm\infty$, and this cannot happen, since $F(x)$ and $G(x)$ are reciprocals (i.e., $G(x)=1/F(x)$). In short, there is no paradox, because L'Hopital does not apply to the ratio $F/G$.