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