Answer
$$\left\| {\nabla w\left( {2,1,1} \right)} \right\| = \sqrt {33} $$
Work Step by Step
$$\eqalign{
& w = x{y^2}{z^2},{\text{ point }}\left( {2,1,1} \right) \cr
& {\text{Find the partial derivatives }}{w_x}{\text{, }}{w_y}{\text{ and }}{w_z}\left( {x,y,z} \right) \cr
& {w_x} = \frac{\partial }{{\partial x}}\left[ {x{y^2}{z^2}} \right] = {y^2}{z^2} \cr
& {w_y} = \frac{\partial }{{\partial y}}\left[ {x{y^2}{z^2}} \right] = 2xy{z^2} \cr
& and \cr
& {w_z} = \frac{\partial }{{\partial z}}\left[ {x{y^2}{z^2}} \right] = 2x{y^2}z \cr
& {\text{Calculate }}\nabla w \cr
& \nabla w = {w_x}{\bf{i}} + {w_y}{\bf{j}} + {w_z}{\bf{k}} \cr
& \nabla w = {y^2}{z^2}{\bf{i}} + 2xy{z^2}{\bf{j}} + 2x{y^2}z{\bf{k}} \cr
& {\text{At the given point the gradient is}} \cr
& \nabla w\left( {2,1,1} \right) = {\left( 1 \right)^2}{\left( 1 \right)^2}{\bf{i}} + 2\left( 2 \right)\left( 1 \right){\left( 1 \right)^2}{\bf{j}} + 2\left( 2 \right){\left( 1 \right)^2}\left( 1 \right){\bf{k}} \cr
& \nabla w\left( {2,1,1} \right) = {\bf{i}} + 4{\bf{j}} + 4{\bf{k}} \cr
& {\text{The maximum value of }}{D_{\bf{u}}}w\left( {x,y,z} \right){\text{ is }}\left\| {\nabla w\left( {x,y,z} \right)} \right\| \cr
& \left\| {\nabla w\left( {2,1,1} \right)} \right\| = \left\| {{\bf{i}} + 4{\bf{j}} + 4{\bf{k}}} \right\| \cr
& \left\| {\nabla w\left( {2,1,1} \right)} \right\| = \sqrt {1 + 16 + 16} \cr
& \left\| {\nabla w\left( {2,1,1} \right)} \right\| = \sqrt {33} \cr} $$
