Answer
Div (curl G) is not zero. Thus, there is no vector field $G$.
Work Step by Step
Given: curl$G=2x i+3y z j-xz^2 k$
Definition of div $F=\nabla \cdot F$
Let us consider $F=pi+qj+rk$
Then, we have div $F=\dfrac{\partial p}{\partial x}+\dfrac{\partial q}{\partial y}+\dfrac{\partial r}{\partial z}$
div (curl G) must be zero when there exists a vector field $G$ for all values of $x,y$ and $z$.
$div (curl G)=\dfrac{\partial (2x)}{\partial x}+\dfrac{\partial (3yz)}{\partial y}+\dfrac{\partial (-xz^2)}{\partial z}=-2+3z-2xz$
It can be seen that div (curl G) is not zero, so there is no such vector field $G$.
Hence, it has been proved.
