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