Answer
$\lt 1+\dfrac{\sqrt 3}{2},\dfrac{\sqrt 3}{2}, \dfrac{-1}{2} \gt $
Work Step by Step
Our aim is to take the first partial derivative of the given function $f(x,y,z)$ with respect to $x$, by treating $y$ and $z$ as a constant, and vice versa:
$f_x=\dfrac{y+1}{\sqrt{1-x^2}}+e^{x+y} \cos z \\ f_y= e^{x+y} \cos z + \tan^{-1} x \\f_z=-e^{x+y} \sin z $
The gradient vector equation is: $\nabla f = \lt f_x,f_y,f_z \gt = \lt e^{x+y} \cos z +\dfrac{y+1}{\sqrt{1-x^2}} , e^{x+y} \cos z +arcsin x , -e^{x+y} \sin z \gt $
Now, $\nabla f (0,0,\dfrac{\pi}{6}) = \lt f_x,f_y,f_z \gt = \lt e^{x+y} \cos z +\dfrac{y+1}{\sqrt{1-x^2}} , e^{x+y} \cos z +\sin^{-1} x , -e^{x+y} \sin z \gt = \lt 1+\dfrac{\sqrt 3}{2},\dfrac{\sqrt 3}{2}, \dfrac{-1}{2} \gt $
