Answer
$${D_u}w\left( {1,2,2} \right) = \frac{2}{3}$$
Work Step by Step
$$\eqalign{
& w = {y^2} + xz,{\text{ }}P\left( {1,2,2} \right),{\text{ }}{\bf{v}} = 2{\bf{i}} - {\bf{j}} + 2{\bf{k}} \cr
& {\text{Calculate }}\nabla w\left( {x,y,z} \right) \cr
& \nabla w\left( {x,y,z} \right) = {w_x}\left( {x,y,z} \right){\bf{i}} + {w_y}\left( {x,y,z} \right){\bf{j}} + {w_z}\left( {x,y,z} \right){\bf{k}} \cr
& {w_x}\left( {x,y,z} \right) = z \cr
& {w_y}\left( {x,y,z} \right) = 2y \cr
& {w_y}\left( {x,y,z} \right) = x \cr
& \nabla w\left( {x,y,z} \right) = z{\bf{i}} + 2y{\bf{j}} + x{\bf{k}} \cr
& {\text{Evaluate }}\nabla w\left( {1,2,2} \right) \cr
& \nabla w\left( {1,2,2} \right) = 2{\bf{i}} + 4{\bf{j}} + {\bf{k}} \cr
& {\text{Calculate }}\left| {\bf{v}} \right| \cr
& \left| {\bf{v}} \right| = \left| {2{\bf{i}} - {\bf{j}} + 2{\bf{k}}} \right| = \sqrt {4 + 1 + 4} = 3 \cr
& {\bf{v}}{\text{ is not a unit vector, the unit vector in the direction of }}{\bf{v}}{\text{ is:}} \cr
& {\bf{u}} = \frac{{\bf{v}}}{{\left| {\bf{v}} \right|}} = \frac{{2{\bf{i}} - {\bf{j}} + 2{\bf{k}}}}{{\sqrt 5 }} = \frac{2}{3}{\bf{i}} - \frac{1}{3}{\bf{j}} + \frac{2}{3}{\bf{k}} \cr
& {\bf{v}}{\text{ is a unit vector.}} \cr
& {\text{The directional derivative at }}\left( {1,2,2} \right){\text{ in the direction of }}{\bf{u}}{\text{ is}} \cr
& {D_u}w\left( {x,y,z} \right) = \nabla w\left( {x,y,z} \right) \cdot {\bf{u}} \cr
& {D_u}w\left( {1,2,2} \right) = \nabla w\left( {1,2,2} \right) \cdot \left( {\frac{2}{3}{\bf{i}} - \frac{1}{3}{\bf{j}} + \frac{2}{3}{\bf{k}}} \right) \cr
& {D_u}w\left( {1,2,2} \right) = \left( {2{\bf{i}} + 4{\bf{j}} + {\bf{k}}} \right) \cdot \left( {\frac{2}{3}{\bf{i}} - \frac{1}{3}{\bf{j}} + \frac{2}{3}{\bf{k}}} \right) \cr
& {D_u}w\left( {1,2,2} \right) = \frac{4}{3} - \frac{4}{3} + \frac{2}{3} \cr
& {D_u}w\left( {1,2,2} \right) = \frac{2}{3} \cr} $$
