Answer
$$0$$
Work Step by Step
Since, $\nabla f =\dfrac{\partial f}{ \partial x} i+\dfrac{\partial f}{ \partial y}j +\dfrac{\partial f}{ \partial z} k$
$ \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, $f=\dfrac{x}{y}+g(y,z)$
or, $\dfrac{\partial g(y,z)}{ \partial x}=0 $
$f=\dfrac{x}{y}+\dfrac{y}{z}+h(z) $
or,$h(z)=C=0$
Now, we will substitute all the above values in the given integral.
$ \int_{1,1,1}^{2,2,2} \dfrac{1}{y} dx +(\dfrac{1}{z} -\dfrac{x}{y^2}) dy -\dfrac{y}{z^2} dz$
or, $=f(2,1,1)-f(1,2,1)=(\dfrac{2}{2}+\dfrac{2}{2}+C)-(\dfrac{1}{1}+\dfrac{1}{1}+C)$
or, $=0$
