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