My version of theorem (to prove):
Let $f:(a,b)\to (c,d)$ be an invertible continuous function, (read carefully) differentiable at $m\in(a,b)$, such that $f'(m)\neq0$. Then $(f^{-1})'(f(m))=\frac1{f'(m)}$.
For convenience, let $g(x)=f^{-1}(x)$.
We have to prove that for every $\epsilon>0$, there exists a $\delta>0$, such that, $$0<|y-f(m)|<\delta \implies|\frac{g(y)-m}{y-f(m)}-\frac{1}{f'(m)}|<\epsilon$$ Now, note that in the above expression, $y\in(c,d)$. Also, for every $y\in(c,d)$ we have a unique $x\in(a,b)$ (and vice versa). ($y=f(x)$)
So above statement can be written as, we have to prove that for every $\epsilon>0$, there exists $\delta>0$, such that, $$0<|f(x)-f(m)|<\delta\implies|\frac{x-m}{f(x)-f(m)}-\frac{1}{f'(m)}|<\epsilon$$
Now since, $f(x)$ is continuous (and injective (think what is the need of one-oneness)), for any positive number $\delta$, we have another number $\delta'$, so that $$0<|x-m|<\delta'\implies 0<|f(x)-f(m)|<\delta$$
So we have to prove that $\forall$ $\epsilon>0$ there exists a $\delta'>0$, so that. $$0<|x-m|<\delta'\implies|\frac{x-m}{f(x)-f(m)}-\frac{1}{f'(m)}|<\epsilon$$ i.e. $$\lim_{x\to m} \frac{x-m}{f(x)-f(m)}=\frac{1}{f'(m)}$$ Which is true by limit theorem.
So is my version of theorem, and the proof, correct?