I feel like I'm missing something obvious here. I understand the graphical proof of $f^{-1'}[a] = {1\over f{'}[f^{-1}(a)]}$, but I'm stuck on one aspect of the simple non-graphical proof. Start with $y = f^{-1}(x)$, and thus $x = f(y)$. If we then take $$f(f^{-1}(x)) = f(y) = x$$ and do implicit differentiation with the chain rule, we get $$\mathrm{f^{'}[y]}\!\cdot\!\mathrm{{dy\over dx}} = 1$$ and the rest follows when you note that ${dy\over dx}$ at $a$ is $f^{-1'}[a]$.
All works out well when you take $f^{'}[y]$ to be ${df\over dx}(y)$. And here's where I get stuck: this is not the derivative that you get when you apply the chain rule (it seems to me, at least). Applying the chain rule to a composition of functions gives the following: $${d\,g(h(x))\over dx} = \mathrm{dg\over \mathbf{dh}}\!\cdot\!\mathrm{dh\over dx}$$ and NOT $${d\,g(h(x))\over dx} = \mathrm{dg\over \mathbf{dx}}\!\cdot\!\mathrm{dh\over dx}$$ So what the differentiation should really give is: $$\mathrm{{df\over dy}}\!\cdot\!\mathrm{{dy\over dx}} = 1,\,\,\,\mathrm{in\,other\,words\!\!:}\,\mathrm{{df\over df^{-1}}}\!\cdot\!\mathrm{{dy\over dx}} = 1$$ This gives a very different result. Example: suppose $f = x^{2}$ so that $y = \sqrt{x} $. On the one hand, ${df\over dx} = 2x$, but on the other hand, ${df\over dy} = \mathrm{df\over dx}\!\cdot\!\mathrm{dx\over dy} = \mathrm{2x}\!\cdot\!\mathrm{2y} = 4x^{{3\over 2}}$. What am I missing?! Thanks in advance.
After you arrive at the following equation $$f'(y)\cdot\frac{dy}{dx} = 1$$
you can just substitute the value of $y$ to arrive at the final answer. This way could save much of your confusion.
What you seem to have done is the following:
When you say that
$$\frac{df}{df^-1}\cdot\frac{dy}{dx} = 1$$
you're absolutely right. But, $f = x$ and $f^-1 = y$. So this equation is essentially the same as
$$\frac{dx}{dy}\cdot\frac{dy}{dx} = 1$$
which is still correct, but all it implies is that
$$\frac{dy}{dx} = \frac{dy}{dx}$$
which is simply an obvious fact, but one that doesn't take you closer to what you're trying to prove.
So, in reality, your math is actually fine.
The mistake is in your example.
In your example, $f^{-1}{'}(x)$ is $\frac{dy}{dx}$ and not $\frac{df}{dy}$
So, $f^{-1}{'}(x) = \frac{dy}{dx} = \frac{d}{dx}x^{\frac{1}{2}} = \frac{1}{2\sqrt{x}}$
You can then easily see that $f'(f^{-1}(x)) = f'(\sqrt(x)) = 2\sqrt(x)$. So the original result you wanted to prove does hold good even in this example.
The confusion is because you've used the same variables to mean different things in your proof and example
Hope it helps!