Answer
$${D_u}w\left( {1,0,1} \right) = \frac{{12}}{{\sqrt 3 }}$$
Work Step by Step
$$\eqalign{
& w = 5{x^2} + 2xy - 3{y^2}z,{\text{ }}P\left( {1,0,1} \right),{\text{ }}{\bf{v}} = {\bf{i}} + {\bf{j}} - {\bf{k}} \cr
& {\text{Calculate }}\nabla w\left( {x,y,z} \right) \cr
& \nabla w\left( {x,y,z} \right) = {w_x}\left( {x,y,z} \right){\bf{i}} + {w_y}\left( {x,y,z} \right){\bf{j}} + {w_z}\left( {x,y,z} \right){\bf{k}} \cr
& {w_x}\left( {x,y,z} \right) = 10{x^2} + 2y \cr
& {w_y}\left( {x,y,z} \right) = 2x - 6yz \cr
& {w_y}\left( {x,y,z} \right) = - 3y \cr
& \nabla w\left( {x,y,z} \right) = \left( {10{x^2} + 2y} \right){\bf{i}} + \left( {2x - 6yz} \right){\bf{j}} - 3y{\bf{k}} \cr
& {\text{Evaluate }}\nabla w\left( {1,0,1} \right) \cr
& \nabla w\left( {1,0,1} \right) = \left( {10 + 0} \right){\bf{i}} + \left( {2 - 0} \right){\bf{j}} - 3\left( 0 \right){\bf{k}} \cr
& \nabla w\left( {1,0,1} \right) = 10{\bf{i}} + 2{\bf{j}} \cr
& {\text{Calculate }}\left| {\bf{v}} \right| \cr
& \left| {\bf{v}} \right| = \left| {{\bf{i}} + {\bf{j}} - {\bf{k}}} \right| = \sqrt {1 + 1 + 1} = \sqrt 3 \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{{{\bf{i}} + {\bf{j}} - {\bf{k}}}}{{\sqrt 3 }} = \frac{1}{{\sqrt 3 }}{\bf{i}} + \frac{1}{{\sqrt 3 }}{\bf{j}} - \frac{1}{{\sqrt 3 }}{\bf{k}} \cr
& {\bf{v}}{\text{ is a unit vector.}} \cr
& {\text{The directional derivative at }}\left( {1,0,1} \right){\text{ in the direction of }}{\bf{u}}{\text{ is}} \cr
& {D_u}w\left( {1,0,1} \right) = \nabla w\left( {x,y,z} \right) \cdot {\bf{u}} \cr
& {D_u}w\left( {1,0,1} \right) = \nabla w\left( {1,0,1} \right) \cdot \left( {\frac{1}{{\sqrt 3 }}{\bf{i}} + \frac{1}{{\sqrt 3 }}{\bf{j}} - \frac{1}{{\sqrt 3 }}{\bf{k}}} \right) \cr
& {D_u}w\left( {1,0,1} \right) = \left( {10{\bf{i}} + 2{\bf{j}} + 0{\bf{k}}} \right) \cdot \left( {\frac{1}{{\sqrt 3 }}{\bf{i}} + \frac{1}{{\sqrt 3 }}{\bf{j}} - \frac{1}{{\sqrt 3 }}{\bf{k}}} \right) \cr
& {D_u}w\left( {1,0,1} \right) = \frac{{10}}{{\sqrt 3 }} + \frac{2}{{\sqrt 3 }} + 0 \cr
& {D_u}w\left( {1,0,1} \right) = \frac{{12}}{{\sqrt 3 }} \cr} $$
