Answer
$$\left\| {\nabla f\left( {2,0, - 4} \right)} \right\| = \sqrt {65} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y,z} \right) = x{e^{yz}},{\text{ point }}\left( {2,0, - 4} \right) \cr
& {\text{Find the partial derivatives }}{f_x}\left( {x,y,z} \right){\text{, }}{f_y}\left( {x,y,z} \right){\text{and }}{f_z}\left( {x,y,z} \right) \cr
& {f_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {x{e^{yz}}} \right] = {e^{yz}} \cr
& {f_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {x{e^{yz}}} \right] = xz{e^{yz}} \cr
& and \cr
& {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {x{e^{yz}}} \right] = xy{e^{yz}} \cr
& {\text{Calculate }}\nabla f\left( {x,y,z} \right) \cr
& \nabla f\left( {x,y,z} \right) = {f_x}\left( {x,y,z} \right){\bf{i}} + {f_y}\left( {x,y,z} \right){\bf{j}} + {f_z}\left( {x,y,z} \right){\bf{k}} \cr
& \nabla f\left( {x,y,z} \right) = {e^{yz}}{\bf{i}} + xz{e^{yz}}{\bf{j}} + xy{e^{yz}}{\bf{k}} \cr
& \nabla f\left( {x,y,z} \right) = {e^{yz}}\left( {{\bf{i}} + xz{\bf{j}} + xy{\bf{k}}} \right) \cr
& {\text{At the given point the gradient is}} \cr
& \nabla f\left( {2,0, - 4} \right) = {e^{\left( 0 \right)\left( { - 4} \right)}}\left( {{\bf{i}} - 8{\bf{j}} + 0{\bf{k}}} \right) \cr
& \nabla f\left( {2,0, - 4} \right) = {\bf{i}} - 8{\bf{j}} \cr
& {\text{The maximum value of }}{D_{\bf{u}}}f\left( {x,y,z} \right){\text{ is }}\left\| {\nabla f\left( {x,y,z} \right)} \right\| \cr
& \left\| {\nabla f\left( {2,0, - 4} \right)} \right\| = \left\| {{\bf{i}} - 8{\bf{j}}} \right\| \cr
& \left\| {\nabla f\left( {2,0, - 4} \right)} \right\| = \sqrt {1 + 64} \cr
& \left\| {\nabla f\left( {2,0, - 4} \right)} \right\| = \sqrt {65} \cr} $$
