Answer
The vector field $F$ is not conservative.
Work Step by Step
$\bf{Definition}:$ If a vector field $F=F_1 i+F_2j$ is a vector field over a simply connected and open set of domain $D$, then the partial derivatives of $ F_1$ and $ F_2$ must satisfy the condition such as: $\dfrac{\partial F_1}{\partial y}=\dfrac{\partial F_2}{\partial x}$
We are given that $F(x,y)=(x^2y, y^2 x)$
$\dfrac{\partial F_1}{\partial y}=x^2 \quad \text{and}\quad
\dfrac{\partial F_2}{\partial x}=y^2$
We see that the partial derivative of a given vector field are not equal. Therefore, the vector field $F$ is not conservative.
