Answer
$$\frac{{x{\bf{i}} + y{\bf{j}} + z{\bf{k}}}}{{\sqrt {{x^2} + {y^2} + {z^2}} }};{\text{ }}1$$
Work Step by Step
$$\eqalign{
& f\left( {x,y,z} \right) = \sqrt {{x^2} + {y^2} + {z^2}} ,{\text{ point }}\left( {1,4,2} \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[ {\sqrt {{x^2} + {y^2} + {z^2}} } \right] = \frac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }} \cr
& {f_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {\sqrt {{x^2} + {y^2} + {z^2}} } \right] = \frac{y}{{\sqrt {{x^2} + {y^2} + {z^2}} }} \cr
& and \cr
& {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {\sqrt {{x^2} + {y^2} + {z^2}} } \right] = \frac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }} \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 = \frac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{i}} + \frac{y}{{\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{j}} + \frac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{k}} \cr
& or \cr
& \nabla f\left( {x,y,z} \right) = \frac{{x{\bf{i}} + y{\bf{j}} + z{\bf{k}}}}{{\sqrt {{x^2} + {y^2} + {z^2}} }} \cr
& \cr
& {\text{At the given point the gradient is}} \cr
& \nabla f\left( {1,4,2} \right) = \frac{1}{{\sqrt {{1^2} + {4^2} + {2^2}} }}\left( {{\bf{i}} + 4{\bf{j}} + 2{\bf{k}}} \right) \cr
& \nabla f\left( {1,4,2} \right) = \frac{1}{{\sqrt {21} }}\left( {{\bf{i}} + 4{\bf{j}} + 2{\bf{k}}} \right) \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( {1,4,2} \right)} \right\| = \left\| {\frac{1}{{\sqrt {21} }}\left( {{\bf{i}} + 4{\bf{j}} + 2{\bf{k}}} \right)} \right\| \cr
& \left\| {\nabla f\left( {1,4,2} \right)} \right\| = \sqrt {{{\left( {\frac{1}{{\sqrt {21} }}} \right)}^2} + {{\left( {\frac{4}{{\sqrt {21} }}} \right)}^2} + {{\left( {\frac{2}{{\sqrt {21} }}} \right)}^2}} \cr
& \left\| {\nabla f\left( {1,4,2} \right)} \right\| = \sqrt {\frac{1}{{21}} + \frac{{16}}{{21}} + \frac{4}{{21}}} \cr
& \left\| {\nabla f\left( {1,4,2} \right)} \right\| = 1 \cr} $$