If ${\tan }^{-1}y-4{\tan }^{-1}x$ then y is not finite if

If we put  $x=\tan \theta$ , the given equality becomes 

$\begin{array}{l}{\tan }^{-1}y=4\theta \\ \Rightarrow y=\tan 4\theta =\frac{2\tan 2\theta }{1-{\tan }^{2}2\theta }=\frac{\left.2{\left[\frac{2\tan \theta }{1-{\tan }^2\theta }\right]}^{\mathrm{\prime }}\right]}{1-{\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)}^2}\end{array}$

$=\frac{2\times 2x\left(1-x^2\right)}{{\left(1-x^2\right)}^2-4x^2}=\frac{4x\left(1-x^2\right)}{1-6x^2+x^4}$

so that y is finite if $x^4-6x^2+1\ne 0$

$\Rightarrow x^2\ne \frac{6\pm \sqrt{36-4}}{2}\ne 3\pm 2\sqrt{2}$