Given function is $y=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-x^{2}}\right)$
Here, domain is $\left[0, \frac{\pi}{4}\right]$
When, $x=0$
$y=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-0}\right) $
When $ x=\frac{\pi}{4}, y=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-\frac{\pi^{2}}{16}}\right) $
$=3 \sin \left(\frac{\pi}{4}\right)=3 \times \frac{1}{\sqrt{2}}=\frac{3}{\sqrt{2}} $
$=3 \sin (0)=0$
Hence, range of given function is $\left[0, \frac{3}{\sqrt{2}}\right]$.