If x, y > 0, then $tan^{-1}\left(\frac{x}{y}\right)-tan^{-1}\left(\frac{x-y}{x+y}\right)$, is 

$\frac{\pi}{4}$

We have,

$tan^{-1}\frac{x}{y}-tan^{-1}\left(\frac{x-y}{x+y}\right)$

$=tan^{-1}\frac{x}{y}-tan^{-1}\left(\frac{1-y/x}{1+y/x}\right)$

$=tan^{-1}\frac{x}{y}-\left(tan^{-1}1 - tan^{-1}\frac{y}{x}\right)$

$= tan^{-1}\frac{x}{y}+tan^{-1}\frac{y}{x}-\frac{\pi}{4}= tan^{-1}\frac{x}{y}+cot^{-1}\frac{x}{y}-\frac{\pi}{4}=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$