Answer
$${D_u}f\left( {1,1,1} \right) = \frac{{2\sqrt 3 }}{3}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2},{\text{ }}P\left( {1,1,1} \right),{\text{ }}{\bf{v}} = \frac{{\sqrt 3 }}{3}{\bf{i}} - \frac{{\sqrt 3 }}{3}{\bf{j}} + \frac{{\sqrt 3 }}{3}{\bf{k}} \cr
& {\text{Calculate }}\nabla f\left( {x,y,z} \right) \cr
& \nabla f\left( {x,y,z} \right) = {f_x}\left( {x,y,z} \right){\bf{i}} + {f_y}\left( {x,y,z} \right){\bf{j}} + {f_z}\left( {x,y,z} \right){\bf{j}} \cr
& {f_x}\left( {x,y,z} \right) = 2x \cr
& {f_y}\left( {x,y,z} \right) = 2y \cr
& {f_z}\left( {x,y,z} \right) = 2z \cr
& \nabla f\left( {x,y,z} \right) = 2x{\bf{i}} + 2y{\bf{j}} + 2z{\bf{j}} \cr
& {\text{Evaluate }}\nabla f\left( {1,1,1} \right) \cr
& \nabla f\left( {1,1,1} \right) = 2{\bf{i}} + 2{\bf{j}} + 2{\bf{j}} \cr
& {\text{Calculate }}\left| {\bf{v}} \right| \cr
& \left| {\bf{v}} \right| = \left| {\frac{{\sqrt 3 }}{3}{\bf{i}} - \frac{{\sqrt 3 }}{3}{\bf{j}} + \frac{{\sqrt 3 }}{3}{\bf{k}}} \right| = \sqrt {\frac{3}{9} + \frac{3}{9} + \frac{3}{9}} = 1 \cr
& {\bf{v}}{\text{ is a unit vector.}} \cr
& {\text{The directional derivative at }}\left( {1,1,1} \right){\text{ in the direction of }}{\bf{u}}{\text{ is}} \cr
& {D_u}f\left( {x,y,z} \right) = \nabla f\left( {x,y,z} \right) \cdot {\bf{v}} \cr
& {D_u}f\left( {1,1,1} \right) = \nabla f\left( {1,1,1} \right) \cdot \left( {\frac{{\sqrt 3 }}{3}{\bf{i}} - \frac{{\sqrt 3 }}{3}{\bf{j}} + \frac{{\sqrt 3 }}{3}{\bf{k}}} \right) \cr
& {D_u}f\left( {1,1,1} \right) = \left( {2{\bf{i}} + 2{\bf{j}} + 2{\bf{j}}} \right) \cdot \left( {\frac{{\sqrt 3 }}{3}{\bf{i}} - \frac{{\sqrt 3 }}{3}{\bf{j}} + \frac{{\sqrt 3 }}{3}{\bf{k}}} \right) \cr
& {D_u}f\left( {1,1,1} \right) = 2\left( {\frac{{\sqrt 3 }}{3}} \right) - 2\left( {\frac{{\sqrt 3 }}{3}} \right) + 2\left( {\frac{{\sqrt 3 }}{3}} \right) \cr
& {D_u}f\left( {1,1,1} \right) = \frac{{2\sqrt 3 }}{3} \cr} $$
