Answer
$\lt -\dfrac{11}{2},-6, \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:
$f_x=-6x+\dfrac{z}{1+(xz)^2} =\dfrac{z}{1+x^2z^2}-6x\\f_y=-6yz \\f_z=6z^2-3x^2-3y^2+\dfrac{x}{1+x^2z^2}$
Write the gradient vector equation.
$\nabla f = \lt f_x,f_y,f_z \gt $
For point $(1,1,1)$
$\nabla f = \lt \dfrac{z}{1+x^2z^2}-6x, -6yz, 6z^2-3x^2-3y^2+\dfrac{x}{1+x^2z^2} \gt = \lt -\dfrac{11}{2},-6, \dfrac{1}{2} \gt $
