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) $f_x=-\sqrt 2 \sin x \sin (y+z) \\ \implies f_x (0,0, \dfrac{\pi}{4})=0$
Next, $f_y=-\sqrt 2 \cos x \cos (y+z) \\ \implies f_x (0,0, \dfrac{\pi}{4})=1$ and $f_z=\sqrt 2 \cos x \cos (y+z) \\ \implies f_x (0,0, \dfrac{\pi}{4})=1$
Now, $L(x,y,z) =1+1(y-0)+(1)(y-0)+1(z-\pi/4)\\ =1+y+z-\dfrac{\pi}{4}$
b) $f_x(\dfrac{\pi}{4},\dfrac{\pi}{4},0)=- \dfrac{\sqrt 2}{2}; \\ f_y(\dfrac{\pi}{4},\dfrac{\pi}{4},0)=\dfrac{\sqrt 2}{2}; \\ f_z(\dfrac{\pi}{4},\dfrac{\pi}{4},0)=\dfrac{\sqrt 2}{2} $
So, $L(\dfrac{\pi}{4},\dfrac{\pi}{4},0) =\dfrac{\sqrt 2}{2}-\dfrac{\sqrt 2}{2}x +\dfrac{\sqrt 2}{2} y +\dfrac{\sqrt 2}{2} z$
