How to prove formally that $x=y²$ is the inverse relation of $y=x²$? I think that , in general, one can find the inverse relation of a function $y = f(x)$ by exchanging $x$'s and $y$'s places, which gives : $x = f(y)$. 
Note : I'm not talking of " inverse function" but simply of " inverse relation". I do not ask for a relation that would necessarily be also a function. 
But how to prove formally that this method is correct? 
In particular, how to prove formally that the inverse relation of 
$\{ (x,y) \mid y = x² \} $
is
$\{ (x,y) \mid x = y² \} $?
How to prove that the ordered pairs belonging to the second relation are the inversed couples of the first relation? 
A desired proof would look like this . It would begin with 
Inverse of R = { (y,x) | y= f(x) }
and end with 
Inverse of R = { (x,y) | x = f(y)} . 
My question is : what syntactical "manoeuvres" allow to exchange x's and y's place on the right of the vertical bar? 
 A: First you need a definition of what an inverse function is.  Then it's almost certain $y=x^2$ fits the definition.
Actually I suppose we also have to know what you mean when you say $x= y^2$ or $y=x^2$ is a relation.
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Definition 1: A relation $R$ on a set $A$ is a subset of $A\times A$..... period.  And we say $aRb$ if $(a,b) \in R$.

When you say $x=y^2$ is a relation you are referring to a set $\{(x,y)| x= y^2\} \subset \mathbb R\times \mathbb R$.  We can call that set/relation $R$. And when we refer to $y=x^2$ as a relation we are referring to the set, $\{(x,y)| y = x^2\}$.  We can call that set $S$.
Now we need a definition of "inverse relation".  I'm not sure what your text uses but I think a definition would be

Definiton 2: $R^{-1}$ is the inverse of $R\subset \mathbb R\times \mathbb R$ if for any $a, b\in A$ then $(a,b) \in R\iff (b,a) \in R^{-1}$....
.. or alternatively.   $R^{-1} = \{(b,a)| (a,b) \in R\}$.

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Now the proof should be straight forward:
If $R = \{(x,y)\in \mathbb R\times \mathbb R|x=y^2\}$ and $S=\{(x,y)\in \mathbb R\times \mathbb R|y=x^2\}$ then
$x=y^2 \iff (x,y) \in \{(x,y) \in \mathbb R\times \mathbb R|x = y^2\} = R$.
$x=y^2 \iff (y,x) \in \{(y,x) \in \mathbb R\times \mathbb R|x = y^2\} =\{(w,z)\in \mathbb R\times \mathbb R|z=w^2\} =\{(x,y)\in \mathbb R\times \mathbb R|y=x^2\}=S$ (variables are nothing but place holders)
So $R^{-1} = S$.
A: It's very simple: the inverse relation of a relation $\mathcal R=\{(x,y\}\subset X\times Y$ is just the relation $\mathcal R^{-1}=\{(y,x)\in Y\times X\mid (x,y)\in \mathcal R\}$.
Therefore, if your relation is $\mathcal R=\{(x, x^2)\mid x\in\mathbf R\}$, the inverse relation is $\mathcal R^{-1}=\{(x^2,x)\mid x\in\mathbf R\}$.
