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How is $\frac{d}{dy}(\frac{dy}{dx}) = \frac{dx}{dy}\frac{d}{dx}(\frac{dy}{dx})$

I was looking at the answer to this question Prove $\frac{d^2y}{dx^2}=-\frac{\frac{d^2x}{dy^2}}{(\frac{dx}{dy})^3}$

I believe this is application of chain rule, but could not intuitively get this. Can I write the above equation as
$\frac{d}{dy}(\frac{dy}{dx}) = \frac{d}{dx}(\frac{dy}{dx})\frac{dx}{dy}$

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    We should first state explicitly what $y$ is. Is it a differentiable function from $\mathbb R$ to $\mathbb R$? – 2017-01-13
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    Yes, let's say $y$ is differentiable function from $ \mathbb R$ to $\mathbb R$ – 2017-01-13
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    What does $\frac{d}{dy} \left( \frac{dy}{dx} \right)$ mean? – 2017-01-13

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According to chain rule, we have, using Leibniz notation

$$\frac{du}{dy}=\frac{dx}{dy}\frac{du}{dx}$$

If $u$ is a function of $y$ and $y$ a function of $x$.

Here, we have $u=\left(\frac{dy}{dx}\right)$, so

$$\frac d{dy}\left(\frac{dy}{dx}\right)=\frac{dx}{dy}\frac d{dx}\left(\frac{dy}{dx}\right)$$

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    Could you explain all these notations please. For the first equality to differentiate wrt $y$ you have to see $u$ as a function of $y$ but then you differentiate $u$ wrt $x$ ? So we see $u(y)=u^*(x(y))$ where $u$ and $u^*$ are different fcts and $x$ is a function of $y$ ? Not to mix the $x$'s, say $u^*=u^*(s)$ where $s$ is a real parameter. Then by chain rule $\frac{du}{dy}(y)=\frac{du^*}{ds}(x(y))\frac{dx}{dy}(y)$ ? In example $u=\frac{dy}{dx}$. Is this $x$ the same from which $x$ is a function of $y$ ? So that we now see $y$ as a function of $x$ ? This is not always possible (IFT) ? – 2017-01-13
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    In the first equation, two different functions are both being called $u$. This is a common abuse of notation when stating the chain rule. I've always thought it's more clear to say: Suppose that $\hat u(y) = u(x(y))$ for all $y \in \mathbb R$. Then $\hat u'(y) = u'(x(y)) x'(y)$ for all $y \in \mathbb R$ (under mild assumptions). But now, $y$ is being used as an independent variable, but also as the name of a function, so I think the notation is quite confusing. – 2017-01-13
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    @littleO Plus the points of evaluations are not even made explicit. Another source of confusion IMO. – 2017-01-13
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    @NeedForHelp I'm sorry if it wasn't very good notation and for not being able to respond sooner. I'm not sure what you mean be this not always being possible in your first comment. If you mean the function may not be differentiable, for example, then please see the comments below the question. If not, please explain :-) – 2017-01-13
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    @Simple Art Sorry, I was asking, not affirmative. I meant, if $y$ is a function of $x$, then $x$ is not necessarily a function of $y$ ? Unless $y(x)$ is bijection? And even so, $y(x)=x^3$ has inverse $x=y^{1/3}$ but $x$ not differentiable at $0$... – 2017-01-13
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    @NeedForHelp ah, I see what you mean. But that is a thing at the level of the question itself, so I suggest you comment that on the main question. – 2017-01-13