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We prove this from the definition of the derivative in terms of limits.

Let\ y=f(x)\\\\ x=f^{-1}(y)\\\\
Let us say x increases by Δx amount and consequently,  y increases by Δy, ie., y becomes (y+Δy).   Thus derivative :

\frac{dy}{dx}= \lim_{\Delta x \to 0} \frac{(y+\Delta y)-y}{(x+\Delta x)-x} =\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}\\

The converse is also true, about the changes in y and x. When y changes from y to (y+Δy) then due to the same relationship between  x and y,  x will change from  x to (x+Δx).  So derivative:

 \frac{dx}{dy}= \lim_{\Delta y \to 0} \frac{(x+\Delta x)-x}{(y+\Delta y)-y} =\lim_{\Delta y \to 0} \frac{\Delta x}{\Delta y}\\\\

Hence the product:
\frac{dy}{dx}*\frac{dx}{dy}=\lim_{\Delta y \to 0} \frac{\Delta y}{\Delta x}*\lim_{\Delta y \to 0} \frac{\Delta x}{\Delta y}\\\\=\lim_{\Delta y \to 0, \Delta x \to 0} \frac{\Delta y}{\Delta x}* \frac{\Delta x}{\Delta y}\\\\=\lim_{\Delta y \to 0, \Delta y \to 0} \frac{\Delta x}{\Delta x}* \frac{\Delta y}{\Delta y}\\\\=\lim_{\Delta y \to 0, \Delta y \to 0}1*1 =1,\ \ As, however\ small\ \Delta x\ is\ \frac{\Delta x}{\Delta x}\ is\ 1\\\\ Hence, \frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}\\ 
Alternately,  the same can be proved using the y = f(x) and y+Δy = f(x+Δx) using the definition of derivatives by  limits.

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