Answer
$$\frac{x}{\sqrt {x^{2}+y^{2}+z^{2}}}i+\frac{y}{\sqrt {x^{2}+y^{2}+z^{2}}}j+\frac{z}{\sqrt {x^{2}+y^{2}+z^{2}}} k$$
Work Step by Step
By using chain rule of differentiation, we have
$f_x(x,y,z)=\frac{x}{\sqrt {x^{2}+y^{2}+z^{2}}}$
and
$f_y(x,y,z)=\frac{y}{\sqrt {x^{2}+y^{2}+z^{2}}}$
and
$f_z(x,y,z)=\frac{z}{\sqrt {x^{2}+y^{2}+z^{2}}}$
Gradient vector field of $f$ is: $$\frac{x}{\sqrt {x^{2}+y^{2}+z^{2}}}i+\frac{y}{\sqrt {x^{2}+y^{2}+z^{2}}}j+\frac{z}{\sqrt {x^{2}+y^{2}+z^{2}}} k$$
