Answer
$\nabla f (x, y, z)=\dfrac{(xi+yj+zk)}{(x^2+y^2+z^2)} $
Work Step by Step
The gradient field of $f(x,y,z)$ can be expressed as:
$\nabla f(x,y,z) =\dfrac{\partial (f(x,y,z))}{\partial x} i+\dfrac{\partial (f(x,y,z))}{\partial y} j+\dfrac{\partial (f(x,y,z))}{\partial z}k $
Since, $f(x,y,z)=\dfrac{1}{2} \ln (x^2+y^2+z^2)$
Therefore,
$\nabla f (x, y, z)=\dfrac{1}{2} [\dfrac{2x}{(x^2+y^2+z^2)}i +\dfrac{2y}{(x^2+y^2+z^2)} j +\dfrac{2z}{(x^2+y^2+z^2)} k ]$
and $\nabla f (x, y, z)=\dfrac{(xi+yj+zk)}{(x^2+y^2+z^2)} $
