#### Answer

$\lt 3,2,-4 \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:
$f_x=2x+\dfrac{z}{x} \\f_y=2y \\f_z= -4z+\ln x$
and $f_y(2,1)=\dfrac{\partial }{\partial x}(y-x)=y^{0}=1|_{(2,1)}$
Write the gradient vector equation.
$\nabla f = \lt f_x,f_y,f_z \gt $
For point $(1,1,1)$
$\nabla f = \lt 2x+\dfrac{z}{x},2y , -4z+\ln x\gt = \lt 2(1)+\dfrac{1}{1},2(1) , -4(1)+\ln (1) \gt = \lt 3,2,-4 \gt $