Answer
$$\nabla w = \frac{1}{{{x^2} + {y^2} + {z^2}}}\left( {x{\bf{i}} + y{\bf{j}} + z{\bf{k}}} \right)$$
Work Step by Step
$$\eqalign{
& w = \ln \sqrt {{x^2} + {y^2} + {z^2}} \cr
& {\text{write the radical }}\sqrt a {\text{ as }}{a^{1/2}}{\text{ and use the logarithmic properties }} \cr
& w = \ln {\left( {{x^2} + {y^2} + {z^2}} \right)^{1/2}} \cr
& w = \frac{1}{2}\ln \left( {{x^2} + {y^2} + {z^2}} \right) \cr
& \cr
& {\text{calculate the partial derivatives }}{w_x}{\text{,}}\,\,{w_y}{\text{, and }}{w_z} \cr
& {w_x} = \frac{\partial }{{\partial x}}\left[ {\frac{1}{2}\ln \left( {{x^2} + {y^2} + {z^2}} \right)} \right] \cr
& {\text{treat }}y{\text{ and }}z{\text{ as constants}} \cr
& {w_x} = \frac{1}{{2\left( {{x^2} + {y^2} + {z^2}} \right)}}\frac{\partial }{{\partial x}}\left[ {{x^2} + {y^2} + {z^2}} \right] \cr
& {w_x} = \frac{1}{{2\left( {{x^2} + {y^2} + {z^2}} \right)}}\left( {2x} \right) \cr
& {w_x} = \frac{x}{{{x^2} + {y^2} + {z^2}}} \cr
& {\text{Similarly}}{\text{, }}\,\,{w_y} = \frac{y}{{{x^2} + {y^2} + {z^2}}}{\text{ and }}{w_z} = \frac{z}{{{x^2} + {y^2} + {z^2}}} \cr
& \cr
& {\text{The gradient of the function }}w\left( {x,y,z} \right){\text{ is defined by }}\left( {{\text{see page 963}}} \right) \cr
& \nabla w = {z_x}{\bf{i}} + {z_y}{\bf{j}} + {z_z}{\bf{k}} \cr
& {\text{substituting the partial derivatives, we obtain}} \cr
& \nabla w = \frac{x}{{{x^2} + {y^2} + {z^2}}}{\bf{i}} + \frac{y}{{{x^2} + {y^2} + {z^2}}}{\bf{j}} + \frac{z}{{{x^2} + {y^2} + {z^2}}}{\bf{k}} \cr
& {\text{factoring}} \cr
& \nabla w = \frac{1}{{{x^2} + {y^2} + {z^2}}}\left( {x{\bf{i}} + y{\bf{j}} + z{\bf{k}}} \right) \cr} $$
