Answer
$${D_u}f\left( {1,2, - 1} \right) = - \frac{1}{{\sqrt 6 }}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y,z} \right) = xy + yz + xz,{\text{ }}P\left( {1,2, - 1} \right),{\text{ }}{\bf{v}} = 2{\bf{i}} + {\bf{j}} - {\bf{k}} \cr
& {\text{Calculate }}\nabla f\left( {x,y,z} \right) \cr
& \nabla f\left( {x,y,z} \right) = {f_x}\left( {x,y,z} \right){\bf{i}} + {f_y}\left( {x,y,z} \right){\bf{j}} + {f_z}\left( {x,y,z} \right){\bf{j}} \cr
& {f_x}\left( {x,y,z} \right) = y + z \cr
& {f_y}\left( {x,y,z} \right) = x + z \cr
& {f_z}\left( {x,y,z} \right) = y + x \cr
& \nabla f\left( {x,y,z} \right) = \left( {y + z} \right){\bf{i}} + \left( {x + z} \right){\bf{j}} + \left( {x + y} \right){\bf{j}} \cr
& {\text{Evaluate }}\nabla f\left( {1,2, - 1} \right) \cr
& \nabla f\left( {1,2, - 1} \right) = {\bf{i}} + 0{\bf{j}} + 3{\bf{j}} \cr
& {\text{Calculate }}\left| {\bf{v}} \right| \cr
& \left| {\bf{v}} \right| = \left| {2{\bf{i}} + {\bf{j}} - {\bf{k}}} \right| = \sqrt {4 + 1 + 1} = \sqrt 6 \cr
& {\bf{v}}{\text{ is not a unit vector, the unit vector in the direction of }}{\bf{v}}{\text{ is:}} \cr
& {\bf{u}} = \frac{{\bf{v}}}{{\left| {\bf{v}} \right|}} = \frac{{2{\bf{i}} + {\bf{j}} - {\bf{k}}}}{{\sqrt 6 }} = \frac{2}{{\sqrt 6 }}{\bf{i}} + \frac{1}{{\sqrt 6 }}{\bf{j}} - \frac{1}{{\sqrt 6 }}{\bf{k}} \cr
& {\text{The directional derivative at }}\left( {1,1,1} \right){\text{ in the direction of }}{\bf{u}}{\text{ is}} \cr
& {D_u}f\left( {x,y,z} \right) = \nabla f\left( {x,y,z} \right) \cdot {\bf{u}} \cr
& {D_u}f\left( {1,2, - 1} \right) = \nabla f\left( {1,2, - 1} \right) \cdot \left( {\frac{2}{{\sqrt 6 }}{\bf{i}} + \frac{1}{{\sqrt 6 }}{\bf{j}} - \frac{1}{{\sqrt 6 }}{\bf{k}}} \right) \cr
& {D_u}f\left( {1,2, - 1} \right) = \left( {{\bf{i}} + 0{\bf{j}} + 3{\bf{j}}} \right) \cdot \left( {\frac{2}{{\sqrt 6 }}{\bf{i}} + \frac{1}{{\sqrt 6 }}{\bf{j}} - \frac{1}{{\sqrt 6 }}{\bf{k}}} \right) \cr
& {D_u}f\left( {1,2, - 1} \right) = \frac{2}{{\sqrt 6 }} + 0 - \frac{3}{{\sqrt 6 }} \cr
& {D_u}f\left( {1,2, - 1} \right) = - \frac{1}{{\sqrt 6 }} \cr} $$
