By the Inverse Function Theorem, this seems obviously, where I would first prove $\tan(x)$ is continuously differentiable on this interval. $\tan(x)$ is continuous and differentiable by it being composed of $2$ continuous/ differentiable functions. Would this suffice? It seems too simple.
Definitions and theorems I can use are: Inverse Function Theorem, Global Inverse Function Theorem, injective, pretty much everything with differentiability and continuity.
I CANNOT use anything dealing with integration. Thank you!
The function $\tan$ restricted to $(-\pi/2,\pi/2)$ is increasing and continuous.
The first property can be proved without differentiation: if $-\pi/2<x<y<\pi/2$, then we have $$ \tan y-\tan x=\frac{\sin y\cos x-\cos y\sin x}{\cos x\cos y}= \frac{\sin(y-x)}{\cos x\cos y}>0 $$ because $0<y-x<\pi$. Thus it is invertible.
It is continuous because it is differentiable, the derivative being $1+\tan^2x$. Note that $1+\tan^2x>0$, so this also proves the function is strictly increasing, hence invertible. By the inverse function theorem, its inverse is everywhere differentiable as well, because the derivative of $\tan$ never vanishes.
The domain of the inverse is $\mathbb{R}$, because of $$ \lim_{x\to-\pi/2^+}\tan x=-\infty \qquad \lim_{x\to\pi/2^-}\tan x=\infty $$ and the intermediate value theorem.
The derivative of the inverse, usually called $\arctan$ can be found from considering $$ \tan\arctan y=y $$ so differentiating both sides with respect to $y$ yields $$ (1+\tan^2\arctan y)\arctan'y=1 $$ and therefore $$ \arctan'y=\frac{1}{1+\tan^2\arctan y}=\frac{1}{1+y^2} $$