If f(x) = ta n ^ - 1 ( ( e ^ 2 / x ^ 3 ) ( e ^ 2 x ^ 3 ) ) + ta n ^...

We have,
$f\left(x\right) = ta{n}^{- 1}\left(\dfrac{2 - 3ln x}{2 + 3ln x}\right) + ta{n}^{- 1}\left(\dfrac{2 + 3ln x}{1 - 6ln x}\right)$
$= ta{n}^{- 1}2 - ta{n}^{- 1}\left(3ln x\right) + ta{n}^{- 1}2 + ta{n}^{- 1}\left(3ln x\right) + \pi$
Using, $\tan^{- 1}A - \tan^{- 1}B = \tan^{- 1}\left(\dfrac{A - B}{1 + AB}\right)$ and

$\tan^{- 1}A + \tan^{- 1}B$$= \tan^{- 1}\left(\dfrac{A + B}{1 - AB}\right) + \pi\left(∵ AB > 1\right)$, we get,
$f\left(x\right) = ta{n}^{- 1}2 + ta{n}^{- 1}\left(2\right) + \pi =$ constant