Answer
$$-\ln 2$$
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}= 2x \ln y -yz$ and $\dfrac{\partial f}{ \partial y}=\dfrac{x^2}{y} -xz$ and $\dfrac{\partial f}{ \partial z}=-xy$
Now, $f(x,y,z)=x^2\ln y -xyz +g(y,z))$
or, $\dfrac{\partial g(y,z)}{ \partial x}=0 $
Now, $f(x,y,z) =x^2\ln y -xyz+h(z)$
and $h(z)=C=0$
Now, we will substitute all the above values in the given integral.
$ \int_{1,2,1}^{2,1,1 } (2 x \ln y ) dx +(\dfrac{x^2}{y} -xz) dy -xy dz$
or, $=f(2,1,1)-f(1,2,1) $
or, $=-\ln 2$
