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