About L'Hopital's rule
Suppose that $f(x)$ and $g(x)$ are differentiable functions at $x = a$ and $f(a) = g(a) = 0$ and $g^{'}(a) \ne 0$.
Then, $\lim\limits_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim\limits_{x \rightarrow a} \frac{f(x) - f(a)}{x - a} \cdot \frac{x - a}{g(x) - g(a)} = \frac{f^{'}(a)}{g^{'}(a)}$.
So in this case, we don't need L'Hopital's rule.
For example, we don't need L'Hopital's rule to calculate $\lim\limits_{x \rightarrow 1} \frac{(2 x - x^4)^\frac{1}{2}-x^\frac{1}{3}}{1-x^{\frac{3}{4}}} = \frac{16}{9}$.
But of course we can apply L'Hopital's rule to calculate $\lim\limits_{x \rightarrow 1} \frac{(2 x - x^4)^\frac{1}{2}-x^\frac{1}{3}}{1-x^{\frac{3}{4}}}$.
Consider the next example:
Suppose $f(x) = x \phi(x)$ and $g(x) = e^x -1$, where $\phi(x)$ is a function which is continuous everywhere but differentiable nowhere.
Then, $f^{'}(0) = \lim\limits_{x \rightarrow 0} \frac{x \phi(x) - 0 \phi(0)}{x - 0} = \lim\limits_{x \rightarrow 0} \phi(x) = \phi(0)$ and $f(x)$ is not differentiable at $x = a \ne 0$. (if $f(x)$ is differentiable at $x = a$, then $\phi(x) = \frac{1}{x} f(x)$ is differentiable at $x = a$. It is a contradiction.)
$f(x)$ and $g(x)$ are differentiable functions at $x = 0$ and $f(0) = g(0) = 0$ and $g^{'}(0) \ne 0$, so $\lim\limits_{x \rightarrow 0} \frac{f(x)}{g(x)} = \frac{f^{'}(0)}{g^{'}(0)} = \frac{\phi(0)}{e^0} = \phi(0)$.
But in this case, we cannot apply L'Hopital's rule to calculate $\lim\limits_{x \rightarrow 0} \frac{f(x)}{g(x)}$ because $f(x)$ is not differentiable on any open interval.
Is the above argument correct?
Yes, in the present discussion, Hôpital's rule only applies when $f'(x)/g'(x)$ has a limit when $x\rightarrow 0$. It is a sufficient condition but not necessary (as your example illustrates).
You might by the way compare to the following proof of l'Hôpital's rule: Suppose $f$ and $g$ are differentiable in a neighborhood $N$ of $0$, that $f(0)=g(0)=0$ and $g(x)\neq 0$ for $x\in N^*=N\setminus\{0\}$ and that $f'(x)/g'(x)$ converges to some constant $c$ as $x$ goes to zero.
Fix any $x\in N^*$ and let $\lambda=\lambda_x=f(x)/g(x)$. Then the function $$ F(t) = f(t)-\lambda g(t)$$ verifies $F(x)=F(0)=0$ so by the MVT there is $t=t_x$ between $x$ and $0$ so that $F'(t_x)=f'(t_x)-\lambda_x g'(t_x)=0$, so $\lambda_x=f'(t_x)/g'(t_x)$. When $x$ goes to $0$ also $t_x$ goes to zero and it follows that $\lambda_x$ converges to $c$ as we wanted to show.