Answer
(a) $L(x,y,z) =1+y+z-\dfrac{\pi}{4}$
(b) $L(\dfrac{\pi}{4},\dfrac{\pi}{4},0) =\dfrac{\sqrt 2}{2}-\dfrac{\sqrt 2}{2}x +\dfrac{\sqrt 2}{2} y +\dfrac{\sqrt 2}{2} z$
Work Step by Step
(a) $f_x (0,0, \dfrac{\pi}{4})=-\sqrt 2 \sin x \sin (y+z) =0 \\ f_y (0,0, \dfrac{\pi}{4})=-\sqrt 2 \cos x \cos (y+z) $
and $f_x (0,0, \dfrac{\pi}{4})=1 \\ f_z(0,0, \dfrac{\pi}{4})=\sqrt 2 \cos x \cos (y+z) $
or, $f_x (0,0, \dfrac{\pi}{4})=1$
and $L(x,y,z) =1+1(y-0)+(1)(y-0)+1(z-\pi/4)\\ =1+y+z-\dfrac{\pi}{4}$
(b) $f_x(\dfrac{\pi}{4},\dfrac{\pi}{4},0)=- \dfrac{\sqrt 2}{2}$
and $f_y(\dfrac{\pi}{4},\dfrac{\pi}{4},0)=\dfrac{\sqrt 2}{2}$
and $ f_z(\dfrac{\pi}{4},\dfrac{\pi}{4},0)=\dfrac{\sqrt 2}{2} $
So, $L(\dfrac{\pi}{4},\dfrac{\pi}{4},0) =\dfrac{\sqrt 2}{2}-\dfrac{\sqrt 2}{2}x +\dfrac{\sqrt 2}{2} y +\dfrac{\sqrt 2}{2} z$
