Answer
$ 9 \ln (2)$
Work Step by Step
Since, $\nabla f =\dfrac{\partial f}{ \partial x} i+\dfrac{\partial f}{ \partial y}j +\dfrac{\partial f}{ \partial z} k$
$\dfrac{\partial f}{ \partial x}= 3x^2$ and $\dfrac{\partial f}{ \partial y}=\dfrac{z^2}{y} $ and $ \dfrac{\partial f}{ \partial z}=2z \ln y$
Now, $f=x^3+g(y,z) $
or, $\dfrac{\partial g(y,z)}{ \partial x}=z^2 \ln y$
$f=x^3 +z^2 \ln y +h(z) $
or, $h(z)=C$
Now, we will substitute all the above values in the given integral.
$ \int_{1,1,1}^{1,2,3} 3x^2 dx +\dfrac{z^2 dy}{y}+(2z) \ln y dz=1+(9) \ln 2 -(1+0)$
or, $= 9 \ln (2)$