$$f(x)=\exp\left(\arctan \frac{x^2}{1-x^2}\right)$$
Find the maximal interval $I$ containing $x = \pi$ on which $f$ is invertible, and find the explicit expression of the inverse function, specifying its domain $D$.
I calculated $f'$ and studied the monotoneity of $f$: this function is increasing for $x>0$, thus $f$ is monotonic, hence injective, hence invertible in $(0,+\infty)$.
So $I=(0,+\infty)$, $D=\left\{y\in \mathbb R: y>0,\tan \log y>-1,\tan\log y\ge0\right\}$ and $$f^{-1}=\sqrt{\frac{\tan(\log y)}{1-\tan(\log y)}}$$
Have I made some mistakes?