Answer
$${D_{\bf{u}}}f\left( {2, - 2,1} \right) = - \frac{3}{{\sqrt 2 }}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y,z} \right) = xy + yz + xz + 4;\,\,\,\,P\left( {2, - 2,1} \right);\,\,\,\,{\bf{u}} = \left\langle {0, - \frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}} \right\rangle \cr
& {\bf{u}} = \sqrt {{{\left( 0 \right)}^2} + {{\left( { - \frac{1}{{\sqrt 2 }}} \right)}^2} + {{\left( { - \frac{1}{{\sqrt 2 }}} \right)}^2}} = 1 \cr
& {\bf{u}}{\text{ is a unit vector}} \cr
& \cr
& {\text{The gradient of }}f\left( {x,y} \right){\text{ is}} \cr
& {f_x}\left( {x,y,z} \right) = y + z \cr
& {f_y}\left( {x,y,z} \right) = x + z \cr
& {f_z}\left( {x,y,z} \right) = x + y \cr
& \nabla f\left( {x,y,z} \right) = \left( {y + z} \right){\bf{i}} + \left( {x + z} \right){\bf{j}} + \left( {x + y} \right){\bf{k}} \cr
& \cr
& {\text{Calculate the gradient at the point }}P\left( {2, - 2,1} \right) \cr
& \nabla f\left( {2, - 2,1} \right) = \left( { - 2 + 1} \right){\bf{i}} + \left( {2 + 1} \right){\bf{j}} + \left( {2 - 2} \right){\bf{k}} \cr
& \nabla f\left( {2, - 2,1} \right) = - {\bf{i}} + 3{\bf{j}} \cr
& \nabla f\left( {2, - 2,1} \right) = \left\langle { - 1,3,0} \right\rangle \cr
& \cr
& {\text{Computing the directional derivatives of }}f{\text{ at }}\left( {2, - 2,1} \right) \cr
& \operatorname{in} {\text{ the direction of the vector }}{\bf{u}} = \left\langle { - 1,3,0} \right\rangle \cr
& {D_{\bf{u}}}f\left( {a,b,c} \right) = \nabla f\left( {a,b,c} \right) \cdot {\bf{u}} \cr
& {D_{\bf{u}}}f\left( {2, - 2,1} \right) = \left\langle { - 1,3,0} \right\rangle \cdot \left\langle {0, - \frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}} \right\rangle \cr
& {D_{\bf{u}}}f\left( {2, - 2,1} \right) = - \frac{3}{{\sqrt 2 }} \cr} $$
