Answer
$3$
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:
$f_x(1,-1,2)=y+z=1 \\ f_y(1,-1,2)=x+z=3 \\f_z(1,-1,2)=x+y=0$
Write the gradient equation.
$\nabla f (1,-1,2)= \lt f_x,f_y, f_z \gt = \lt 1,3,0 \gt$
Thus, $v=\dfrac{u}{|u|} = \dfrac{\lt 3,6, -2 \gt }{\sqrt {(3)^2+(6)^2+(-2)^2}} =\lt \dfrac{3}{7}, \dfrac{6}{7}, \dfrac{-2}{7}\gt$
The directional derivative at that direction is given as:
$D_v f=\lt 1,3,0 \gt \times \lt \dfrac{3}{7}, \dfrac{6}{7}, \dfrac{-2}{7}\gt=3$
