Answer
$\nabla g(x,y,z)=(y+z) i +(x+z) j +(y+x) k$
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, $ g(x,y,z)=xy+yz+xz$
$\nabla f(x,y,z) =\dfrac{\partial (f(x,y,z))}{\partial x} i+\dfrac{\partial (f(x,y,z))}{\partial y} j+\dfrac{\partial (f(x,y,z))}{\partial z}k
\\ =\dfrac{\partial (f(x,y,z))}{\partial x} i+\dfrac{\partial (f(x,y,z))}{\partial y} j+\dfrac{\partial (f(x,y,z))}{\partial z}k \\=(y+z) i +(x+z) j +(y+x) k$
and $\nabla g(x,y,z)=(y+z) i +(x+z) j +(y+x) k$
