$\text{ The range of the function }f\left(x\right)=\tan \sqrt{\frac{{\pi }^2}{9}-x^2}\text{ is }$

$\text{ For }f\left(x\right)\text{ to be defined },\frac{{\pi }^2}{9}-x^2\ge 0$

$\begin{array}{r}\Rightarrow -\frac{\pi }{3}\le x\le \frac{\pi }{3}\\ \text{ Domain of }f=\left[-\frac{\pi }{3},\frac{\pi }{3}\right]\end{array}$

The greatest value of
$\begin{array}{r}f\left(x\right)=\tan \sqrt{\frac{{\pi }^2}{9}-0}=\sqrt{3 }\text{ when }x=0\\ \end{array}$

and the least value of 

$\begin{array}{r}f\left(x\right)=\tan \sqrt{\frac{{\pi }^2}{9}-\frac{{\pi }^2}{9}}=0\text{ when }x=\frac{\pi }{3}\\ \end{array}$

$\text{ Range of function }=[0,\sqrt{3}]$