find range of f(x)=3sin(root(pi²/16) -x²)

the given function is $f\left(x\right)=3.\sin \left(\sqrt{\frac{{\pi }^2}{16}-x^2}\right)$
since the quantity within the square root can not be negative.
therefore
$$\begin{gathered} \frac{{\pi }^2}{16}-x^2\ge 0 \\ {x}^2\le \frac{{\pi }^2}{16} \\ -\frac{\pi }{4}\le x\le \frac{\pi }{4} \end{gathered}$$
the minimum value of the function is $3.\sin 0=0$
and the maximum value of the function is $3\sin \frac{\pi }{4}=\frac{3}{\sqrt{2}}$

therefore range is $\left[0,\frac{3}{\sqrt{2}}\right]$
hope this helps you