Answer
$ f_{z}(x, y, z)=\displaystyle \frac{\sin z\cos z}{\sqrt{\sin^{2}x+\sin^{2}y+\sin^{2}z}},\quad f_{z} (0,0,\ \displaystyle \frac{\pi}{4})= \displaystyle \frac{\sqrt{2}}{2}$
Work Step by Step
$f(x, y, z) =\sqrt{\sin^{2}x+\sin^{2}y+\sin^{2}z}=(\sin^{2}x+\sin^{2}y+\sin^{2}z)^{1/2}$
Treat x and z as constant to calculate $f_{z}(x, y ,z)$.
Apply the chain rule (twice).
$f_{z}(x, y, z)=\displaystyle \frac{1}{2}(\sin^{2}x+\sin^{2}y +\displaystyle \sin^{2}z)^{-1/2}[\frac{\partial}{\partial z}(\sin^{2}x+\sin^{2}y+\sin^{2}z)]$
$=\displaystyle \frac{1}{2}(\sin^{2}x+\sin^{2}y +\sin^{2}z)^{-1/2}(0+0+2\sin z \cos z)$
$=\displaystyle \frac{\sin z\cos z}{\sqrt{\sin^{2}x+\sin^{2}y+\sin^{2}z}}$
$f_{z} (0,0,\ \displaystyle \frac{\pi}{4})=\frac{\sin\frac{\pi}{4}\cos\frac{\pi}{4}}{\sqrt{\sin^{2}0+\sin^{2}0+\sin^{2}\frac{\pi}{4}}}$
$=\displaystyle \frac{\frac{\sqrt{2}}{2}.\frac{\sqrt{2}}{2}}{\sqrt{0+0+(\frac{\sqrt{2}}{2})^{2}}}$
$=\displaystyle \frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}}$
$=\displaystyle \frac{1}{\sqrt{2}} = \displaystyle \frac{\sqrt{2}}{2}$.
