Answer
$0$
Work Step by Step
Since, $\nabla f =\dfrac{\partial f}{ \partial x} i+\dfrac{\partial f}{ \partial y}j +\dfrac{\partial f}{ \partial z} k ...(1)$
We have: $ \dfrac{\partial f}{ \partial x}= \dfrac{1}{y} ; \dfrac{\partial f}{ \partial y}=\dfrac{-x}{y^2} +\dfrac{1}{z} \\ \dfrac{\partial f}{ \partial z}=\dfrac{-y}{z^2}$
Now, $f=\dfrac{x}{y}+g(y,z) ...(2)$
$\implies \dfrac{\partial g(y,z)}{ \partial x}=0 $
Equation (2) becomes: $f=\dfrac{x}{y}+\dfrac{y}{z}+h(z) .....(3)$
and $h(z)=C=0$
Substitute all the above values in the given integral as follows:
$ \int_{1,1,1}^{2,2,2} \dfrac{1}{y} dx +(\dfrac{1}{z} -\dfrac{x}{y^2}) dy -\dfrac{y}{z^2} dz=f(2,1,1)-f(1,2,1)=(\dfrac{2}{2}+\dfrac{2}{2}+C)-(\dfrac{1}{1}+\dfrac{1}{1}+C) \\ =0$
