Why is this theorem about derivatives true? $\frac{dy}{dx}= \frac{1}{dx/dy}$ 
$$\frac{dy}{dx}= \frac{1}{\;\frac{dx}{dy}\;}$$
Why is the above theorem true as long as $dx/dy$ is not zero? How can you prove it rigorously?

I don’t think it is obvious by the definition of the derivative. I think this says $dx/d(x^2)$ will equal to $1/2x$ and so we can evaluate derivatives such as this. But I want a rigorous proof.
Edit: by the answers I think you want the existence and differentiability of f inverse for something like this to even work ? Could the derivative still exist in such an example and fail to be able to be evaluated like this?Or does that have no meaning ?
 A: This question has some good answers already, but I want to point out that the intuition from abusing the notation can lead to a proof directly.
Just using the limits behind the derivative notation works:
$$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta y \to 0} \frac{\Delta y}{\Delta x} =\lim_{\Delta y \to 0} \frac{1}{\Delta x / \Delta y} = \frac{1}{\lim_{\Delta y \to 0}\Delta x / \Delta y} = \frac{1}{dx/dy}$$

But this does require some extra explaination. First of all, we assume
$y = f(x), x = f^{-1}(y)$, i.e. $y$ is a function of $x$ and vice versa. We need that these functions are differentiable so that all limits written above exist. Next
$$\Delta x = x_2 - x_1,\quad \Delta y = f(x_2) - f(x_1) = y_2 - y_1$$
or equivalently
$$\Delta y = y_2 - y_1,\quad \Delta x = f^{-1}(y_2) - f^{-1}(y_1) = x_2 - x_1$$
Last but not least, because $f$ and $f^{-1}$ are continuous (because they are differentiable), we have that
$$\Delta x \to 0\iff\Delta y \to 0$$
which I used in the beginning.
A: Let $y=f(x)$ with  the inverse function of $x=g(y)$.
We have $$f(g(y))=y$$
Apply the chain rule to get$$ f'(g(y))g'(y) =1$$ Thus $$ f'(g(y))=\frac {1}{g'(y)}$$
That is $$\frac {dy}{dx}= \frac {1}{\frac {dx}{dy}}$$
A: If
$$
f^{-1}(f(x)) = x
$$
in some neighborhood of $x$, then by the chain rule,
$$
\dfrac{df^{-1}(f(x))}{dy} f'(x)= 1,
$$
and
$$
\dfrac{df^{-1}(y)}{dy}= \dfrac{1}{f'(x)}
$$
where $y = f(x)$.
A: Consider the chain rule.
But first off note, neither $\frac {dy}{dx}$ nor $\frac {dx}{dy}$ need not exist or make any sense.  But if they do there there exist some function $f$ where:
$y = f(x)$ and $f$ is differentiable and $f$ is invertable so $x = f^{-1}(y)$ and $f^{-1}$ is differentiable.  And we have that $x = f^{-1}(f(x))$ and $y = f(f^{-1}(y))$.
If we accept that is our premise we can just use the chain rule.
On the one hand we have the identity function $i(x) = x$ and $i'(x) =1$ or in Leibniz notation $\frac {dx}{dx} = 1$.
But if we view $i(x)$ as a composite function $i(x) = f^{-1}(f(x))$ then we can derive the derivative via the chain rule: we have $i'(x)=[f^{-1}]'(f(x))\cdot f'(x)=[f^{-1}]'(y)\cdot f'(x)$ or in Leibniz notation $\frac {dx}{dx} =\frac {d(f^{-1}(f(x))}{dx} = \frac {d(f^{-1}(f(x))}{d(f(x))}\frac {d(f(x))}{dx}=\frac {dx}{dy}\frac {dy}{dx}$
But bearing in mind that $i'(x) = 1$ or $\frac {dx}{dx} = 1$ we just manipulate:
$i'(x)=[f^{-1}]'(f(x))\cdot f'(x)=[f^{-1}]'(y)\cdot f'(x)=1$ so $f'(x)=\frac 1{[f^{-1}]'(f(x))}= \frac 1{[f^{-1}]'(y)}$.  Or in Leibniz notation $\frac {dx}{dx} =\frac {d(f^{-1}(f(x))}{dx} = \frac {d(f^{-1}(f(x))}{d(f(x))}\frac {d(f(x))}{dx}=\frac {dx}{dy}\frac {dy}{dx}=1$ so $\frac {dy}{dx} = \frac 1{\frac {dx}{dy}}$.
====
We could prove this directly similar to how we prove the chain rule:
$1 = \lim_{x\to a} \frac {x-a}{x-a} =\lim_{x\to a}\frac {f^{-1}(f(x))- f^{-1}(f(a))}{x-a}=$
$\lim_{x\to a} \frac {f^{-1}(f(x))- f^{-1}(f(a))}{f(x) - f(a)}\frac {f(x)-f(a)}{x-a}=$  (assuming $f$ is continuous and ... nice)
$\lim_{f(x)\to f(a)}\frac {f^{-1}(f(x))- f^{-1}(f(a))}{f(x) - f(a)}\lim_{x\to a} \frac {f(x)-f(a)}{x-a}=$
$[f^{-1}]'(f(a))\cdot f'(a)$.
A: 
I think this says dx/d(x2) will equal to 1/2x and so we can evaluate derivatives such as this. But I want a rigorous proof.

If $y= x^2$ then $x =\begin{cases}\sqrt y&x\ge0\\-\sqrt y& x \le 0\end{cases}$
If $x \ge 0$ then $\frac {dx}{d(x^2)} = \frac {d\sqrt{y}}{dy}=\frac 1{2\sqrt{y}} = \frac 1{2x}$.
If $x \le 0$ then $\frac {dx}{d(x^2)} = \frac {d(-\sqrt{y})}{dy}= -\frac 1{2\sqrt{y}} = \frac 1{2x}$.
So $\frac {dx}{d(x^2)} = \frac 1{2x}$.
......
Provided there is an $f$ so that $y = f(x)$ and $x = f^{-1}(y)$. We can always have
$1 = \frac {dx}{dx}= \frac {dx}{df(x)}\frac {df(x)}{dx}=\frac {df^{-1}(f(x))}{df(x)}\frac {df(x)}{dx}= \frac {df^{-1}(y)}{dy}\frac {dy}{dx} =\frac {dy}{dx}\frac {dx}{dy}$
A: It can be shown quite easily that if $f:[a,b]\rightarrow [c,d] $ is differentiable some $x_0 \in [a,b]$ then there’s a function $\Phi $ that is continuous at $x_0 $ and $\Phi (x_0)=f’(x_0) $ and $$f(x)=f(x_0)+ \Phi (x)(x-x_0) $$. The converse also holds.
Suppose we have a continuous bijection $f:[a,b]\rightarrow [c,d] $ is differentiable at $x_0$ and $f’(x_0)\neq 0 $.
Let $y=f(x)$ and $y_0=f(x_0) .$
Then there is a function $\Phi $ continuous at $x_0$ such that $\Phi (x_0)=f’(x_0)\neq 0 $ and $$f(x)=f(x_0) +\Phi (x)(x-x_0). $$
Now since $\Phi $ is continuous at $x_0$ then $\Phi (x)\neq 0$ close enough to $x_0$. So $1/\Phi $ is defined close enough to $x_0$.
Now we have $$f^{-1}(y)=f^{-1} (y_0) +(1/\Phi )(f^{-1}(y))(y-y_0).$$
We know $\Phi f^{-1} $ is continuous at $y_0$ since $f^{-1}$ continuous at $y_0$ and $\Phi $ continuous at $x_0=f^{-1}(y_0)$. So $f^{-1} $ is differentiable at $y_0 $ and $$(f^{-1})’(y_0)=(1/\Phi (f^{-1}(y_0)) = \frac{1}{f’(x_0)}.$$
A: I think it is better to understand the result with some more clarity.
Let's first begin with the continuity part:

Theorem 1: Let a function $f:[a, b] \to\mathbb {R} $ be strictly monotone and continuous on $[a, b] $ and let $I=f([a, b]) $ be the range of $f$. Then there exists a function $g:I \to\mathbb {R} $ such that $g$ is continuous on $I$ and $$f(g(x)) =x\, \forall x\in I, g(f(x)) =x\, \forall x\in[a, b] $$

The function $g$ is unique and traditionally denoted by $f^{-1}$ and the important point in the above theorem is that inverse of a continuous function is also continuous. Also observe that if a continuous function is invertible it must also be one-one and continuity combined with one-one nature on an interval forces the function to be strictly monotone. Another point worth remarking is that $I=f([a, b]) $ is also an interval which is either $[f(a), f(b)] $ or $[f(b), f(a)] $ depending upon whether $f$ is increasing or decreasing.
You should be able to prove the above theorem using properties of continuous functions on a closed interval.
Once we are done with the continuity part it is not much difficult to deal with derivatives and we have:

Theorem 2: Let a function $f:[a, b] \to\mathbb {R} $ be strictly monotone and continuous on $[a, b] $. Let $c\in (a, b) $ be such that $f'(c) \neq 0$ and $d=f(c) $. Then the inverse function $f^{-1}$ is differentiable at $d$ with the derivative given by $$(f^{-1})'(d)=\frac{1}{f'(c)}=\frac{1}{f'(f^{-1}(d))}$$

Before coming to the proof of the theorem above it is best to illustrate it via a typical example. So let $f:[-\pi/2,\pi/2]\to\mathbb{R}$ be defined by $f(x) =\sin x$ and the range of $f$ here is $I=[-1,1]$. The derivative $f'(x) =\cos x$ is non-zero in $(-\pi/2,\pi/2)$ and hence the inverse function $f^{-1}$ (usually denoted by $\arcsin$) is differentiable on $(-1,1) $.
To evaluate $(f^{-1})'(x)$ for $x\in (-1,1)$ we need to use a point $y\in(-\pi/2,\pi/2)$ such that $x=f(y) =\sin y$ and  we have $$(f^{-1})'(x)=\frac{1}{f'(y)}=\frac{1}{\cos y}=\frac{1}{\sqrt{1-\sin^2y}}=\frac{1}{\sqrt{1-x^2}}$$ The proof of above theorem is based on the definition of derivative. One should note that as per theorem 1 the inverse function $f^{-1}$ is continuous on range of $f$ and in particular at point $d=f(c) $. We have
\begin{align}
(f^{-1})'(d)&=\lim_{h\to 0}\frac{f^{-1}(d+h)-f^{-1}(d)}{h}\notag\\
&=\lim_{k\to 0}\frac{k}{f(c+k)-f(c)}\notag\\
&=\frac{1}{f'(c)}\notag
\end{align}
Here we have used $$k=f^{-1}(d+h)-f^{-1}(d)=f^{-1}(d+h)-c$$ so that $$d+h=f(c+k)$$ or $$h=f(c+k) - d=f(c+k) - f(c) $$ and note that by continuity of $f^{-1}$ at $d$ we have $k\neq 0,k\to 0$ as $h\to 0$.
It should be observed that for the result to hold we must ensure that derivative $f'(c) \neq 0$ and $f^{-1}$ is continuous at $d=f(c) $.
A: The theorem you ask for is true in a completely rigorous generalized framework for derivatives, as described in this post with the desired theorem and the proof. For convenience I shall reproduce the two examples from that post, which show without doubt that conditions often claimed to be needed are in fact not. Obviously, if you choose to work in a limited framework, you may not be able to get the same generalized results, but that merely reveals a limitation of the chosen framework rather than a limitation of the theorem itself. For the precise definitions please refer to the linked post.
First is a function that is differentiable only at $0$ but has a discontinuity in every open interval around $0$. Same goes for its inverse.

Second is a curve that has a well-defined derivative when it passes through the origin but is not locally bijective there:

