Proving that $\tan{x}-x$ has an inverse and than determining it Consider this problem;
Show that the function $\tan{x}-x$ can be inverted and find the corresponding inverse for which applies
$$ f^{-1} (y) = (f^{-1}(y) + y)^{-2}$$
So to prove that a function can be inverted it has to be bijective, which means it has to be surjective and injective
Now to prove for injective I derived the function $\tan{x}$ and got that it is constantly growing so it is injective. To prove that it is surjective we are already given a solution. Now what I did is derived the function $\tan{x}-x$ and got that its simply $\tan^2{x}$ since the $1$ and $-1$ cancel out. Then out of the starting equation I found a way to explicitly write $\tan{x}$;
$$\tan{x}=f(x)+x$$ and with that we can also determine this;
$$\tan^2{x}=(f(x)+x)^2$$
Now I've used the formula $$ (f^{-1})'(y) = \frac {1}{f'(x)} $$ and this is where I am stuck, the only progress I could make is if instead of $f'$ I insert $\tan^2{x}$ but that is it.
Looking forward to your answers !
EDIT: The domain of the function was also given (my bad)
$$\left(\mbox{-}\frac{\pi}{2}, \frac{\pi}{2}\right)$$
 A: Here are 2 closed form solutions, one with a limit, for the inverse of $\tan(x)-x$ using the Bessel J Zero and Inverse Beta Regularized functions:
$$\tan(y)-y=x\implies x=\tan^{-1}\left(\text j_{\frac32,n+\frac x\pi}\right)+n\pi$$
which works and $n$ gives the $n$th root in terms of increasing size.
Similarly from:

Inverse of $x - \tanh(x)$:

$$x\in\text{Imaginary}\ \Bbb I:\tan(y)-y= x\implies y=\pm i\lim_{a\to0}\tanh^{-1}\left(\sqrt{\text I^{-1}_{\pm aix}\left(\frac32,\frac a2\right)}\right)\sim  1-ix$$
which works. Note the $2$ symmetrical “branches” of the inverse where both “$+$”s or “$-$”s are taken based on the sign of $x$. Please correct me and give me feedback!
This inverse is probably the only one that can be done with a Bessel Zero and a quantile function in Mathematica.
A: COMMENT.-There is no difficulty in verifying that the function $f (x) = \tan (x) -x$ is a bijection of $(-\dfrac{\pi}{2},\dfrac{\pi}{2})$ on $\mathbb R$. Furthermore, as many points of the inverse function can be constructed in the Cartesian plane as desired. Since if $(x, f (x))$ is a point on the initial curve then the symmetric with respect to the first diagonal, $(f (x), x)$ is a point of $f^{-1}$. In the attached figure, the red points of the inverse function have been drawn as symmetrical to the green points on the curve $f (x) = \tan (x) -x$ and the curve corresponding to $f^{-1}$ has been drawn with imperfect drawing in blue color.
I don't think there is a simple expression for the inverse function. But you can find, as I repeat, as many points of the graph of $f^{-1}$ as you want.

A: Okay so as I've just mentioned in the comments I've been unable to solve this so I looked in the solutions.The solution is very unclear to me,so I'd like if someone could try to explain or give me a step in the right direction.This is what the solution in its full detail looks.
$$ f(x) = tan x - x => tan(x) = f(x) + x$$
$$ f'(x) = 1 + tan^2(x) - 1 = tan^2(x) $$
$$ tan^2(x) = (f(x) + x)^2 $$
$$ (f^-1)'(x) = \frac{1}{f'(f^-1)} = \frac{1}{tan^2(f^-1)} = \frac{1}{(f(f^-1)+f^-1)^2} =\frac{1}{(x+f^-1(x))^2}$$
$$ (f^-1)'(y) = \frac{1}{f'(f^-1(y))}$$
Now the only thing more standing in the solutions is that $$(f(f^-1) = id(x) = x $$
Could anyone give me insight on what happened here?
Thank you.
