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