Answer
$ y $ is continuous on
$$... \cup(-\frac{\pi}{2},\frac{\pi}{2})\cup(\frac{\pi}{2},\frac{3\pi}{2}) \cup(\frac{3\pi}{2},\frac{5\pi}{2})\cup...$$
Work Step by Step
Given $$ y=\frac{x\tan x}{x^2+1} =\frac{x}{x^2+1}\frac{\sin x}{\cos x}$$
Since $ x^2+1 \neq 0 \ \ \forall x \in R $, the denominator is zero at:
$\cos x=0 \Rightarrow x= \frac{\pi }{2}+k\pi \Rightarrow x=2k+1, \ \ k \in Z $
So, $ y $ is continuous on
$$... \cup(-\frac{\pi}{2},\frac{\pi}{2})\cup(\frac{\pi}{2},\frac{3\pi}{2}) \cup(\frac{3\pi}{2},\frac{5\pi}{2})\cup...$$
