Answer
$\lt 1+\dfrac{\sqrt 3}{2},\dfrac{\sqrt 3}{2}, \dfrac{-1}{2} \gt $
Work Step by Step
In order to find the partial derivative, we will differentiate
with respect to $x$, by keeping $y$ as a constant to find the x-coordinate of the gradient vector, and vice versa:
$f_x=e^{x+y} \cos z +\dfrac{y+1}{\sqrt{1-x^2}} \\ f_y= e^{x+y} \cos z +arcsin x \\f_z=-e^{x+y} \sin z $
Write the gradient vector equation.
$\nabla f = \lt f_x,f_y,f_z \gt $
$\implies \nabla f = \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 $
Thus, $\nabla f (0,0,\dfrac{\pi}{6}) = \lt 1+\dfrac{\sqrt 3}{2},\dfrac{\sqrt 3}{2}, \dfrac{-1}{2} \gt $
