Implication in L'Hôpital's Rule When an unknown quotient $ \dfrac{0}{0} $ is indeterminate  it is hypothesized as an indeterminate maximum /minimum   point, is n't it?
For, when quotient of two functions
$ \dfrac{x}{y} = \dfrac{0}{0} = const.,$  is considered and we apply Quotient Rule for the first indeterminate fraction by differentiating w.r.t. $t$,
$$  \dfrac {y x^{\prime} - y^{\prime}  x}{y^2} = 0,  \quad  \dfrac {x}{y} =   \dfrac {x^{\prime} }{y^{\prime} }  \tag{1} $$
is what forms the L'Hôpital's Rule as a consequence. Right?
Now how far is the assumption of the unknown to be locally constant, and not be a variable,correct?
EDIT
Apologies. An error occurred while considering multi-variable calculus context, but here makes no sense.Not deleting it as there are already some answers. Sincere apology again for attentions gone improperly, prefer to close it in its present form.

One possible way to consider the problem : assuming that functions $f(x)$ and $g(x)$ have all required properties around $x=a$, use Taylor expansions $$f(x)=f(a)+(x-a) f'(a)+\frac{1}{2} (x-a)^2 f''(a)+\frac{1}{6} f'''(a)
   (x-a)^3+O\left((x-a)^4\right)$$ $$g(x)=g(a)+(x-a) g'(a)+\frac{1}{2} (x-a)^2 g''(a)+\frac{1}{6} g'''(a)
   (x-a)^3+O\left((x-a)^4\right)$$ and now, consider what happens to $$\frac{f(x)}{g(x)}=\frac{f(a)+(x-a) f'(a)+\frac{1}{2} (x-a)^2 f''(a)+\frac{1}{6} f'''(a)
   (x-a)^3+O\left((x-a)^4\right) }{ g(a)+(x-a) g'(a)+\frac{1}{2} (x-a)^2 g''(a)+\frac{1}{6} g'''(a)
   (x-a)^3+O\left((x-a)^4\right)}$$ when $f^{(n)}(a)$ and/or $g^{(m)}(a)$ tend to $0$.
