Answer
$f$ is discontinuous at $x=\left[(2n+1)\displaystyle\frac{\pi}{2}\right]^2$, where $n$ is integer.
Work Step by Step
\[y=f(x)=\tan\sqrt{x}\]
\[\Rightarrow f(x)=\frac{\sin\sqrt x}{\cos \sqrt x}\]
Clearly $f$ is continuous at all values of $x$ for which denominator of $f$ is 0.
\[\cos\sqrt{x}=0\Rightarrow \sqrt{x}=(2n+1)\frac{\pi}{2}\Rightarrow x=\left[(2n+1)\frac{\pi}{2}\right]^2\]
$\Rightarrow f$ is discontinuous at $x=\left[(2n+1)\frac{\pi}{2}\right]^2$, where $n$ is integer.
