Answer
$$\left\| {\nabla z\left( {1,1} \right)} \right\| = 4$$
Work Step by Step
$$\eqalign{
& z = \frac{{{x^2}}}{{x - y}},{\text{ point }}\left( {2,1} \right) \cr
& {\text{Find the partial derivatives }}{z_x}{\text{ and }}{z_y} \cr
& {z_x} = \frac{\partial }{{\partial x}}\left[ {\frac{{{x^2}}}{{x - y}}} \right] \cr
& {z_x} = \frac{{2x\left( {x - y} \right) - {x^2}}}{{{{\left( {x - y} \right)}^2}}} \cr
& {z_x} = \frac{{{x^2} - 2xy}}{{{{\left( {x - y} \right)}^2}}} \cr
& and \cr
& {z_y} = \frac{\partial }{{\partial y}}\left[ {\frac{{{x^2}}}{{x - y}}} \right] \cr
& {z_y} = \frac{{{x^2}}}{{{{\left( {x - y} \right)}^2}}} \cr
& {\text{Calculate }}\nabla z \cr
& \nabla z = {z_x}{\bf{i}} + {z_y}{\bf{j}} \cr
& \nabla z = \frac{{{x^2} - 2xy}}{{{{\left( {x - y} \right)}^2}}}{\bf{i}} + \frac{{{x^2}}}{{{{\left( {x - y} \right)}^2}}}{\bf{j}} \cr
& {\text{At the given point the gradient is}} \cr
& \nabla z\left( {2,1} \right) = \frac{{{2^2} - 4}}{{{{\left( 1 \right)}^2}}}{\bf{i}} + \frac{{{2^2}}}{{{{\left( 1 \right)}^2}}}{\bf{j}} \cr
& \nabla z\left( {2,1} \right) = 0{\bf{i}} + 4{\bf{j}} \cr
& {\text{The maximum value of }}z\left( {x,y} \right){\text{ is }}\left\| {z\left( {x,y} \right)} \right\| \cr
& \left\| {\nabla z\left( {1,1} \right)} \right\| = \left\| {4{\bf{j}}} \right\| \cr
& \left\| {\nabla z\left( {1,1} \right)} \right\| = 4 \cr} $$
