If $\,\,y=f^{-1}(x)$,
Then $x=f(y)$.
So $\,\,\frac{dx}{dy}=f'(x)$,
And $\frac{dy}{dx}=\frac{1}{f'(x)}$.
Using the principle of differentiating by first principles,
Then $$\frac{dx}{dy}=\lim_{h\to 0}\,\,\frac{x(y+h)-x(y)}{h},$$
And, $$\frac{dy}{dx}=\lim_{h \to 0}\,\,\frac{y(x+h)-y(x)}{h}.$$
Does this mean that, $$\begin{equation}\lim_{h\to 0}\,\,\frac{x(y+h)-x(y)}{h}=\left[ \lim_{h \to 0}\,\,\frac{y(x+h)-y(x)}{h} \right] ^{-1} \end{equation},\qquad(1)$$
Or $$\lim_{h\to 0}\,\,\frac{x(y+h)-x(y)}{h}= \lim_{h \to 0}\,\,\left[\frac{y(x+h)-y(x)}{h} \right] ^{-1}?\qquad(2)$$
Or do both statements $(1)$ and $(2)$ mean essentially the same thing?