Answer
The vector field $\overrightarrow{F}$ is not conservative.
Work Step by Step
The vector field $F(x,y)=Pi+Qj$ is a conservative field throughout the domain $D$, when
$\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$
Here, $P$ and $Q$ are the first-order partial derivatives on the domain $D$.
$\int_C \overrightarrow{F} \cdot \overrightarrow{dr}$ is independent of the path if and only if the line integral $\int_C \overrightarrow{F} \cdot \overrightarrow{dr}=0$ for every closed curve $C$.
We know that the work done is the line integral of force. When we draw a closed loop around the center of the vector field, then we get a non-zero amount of work needed to get to the initial point.
For a conservative field, there will always be an equal amount of positive and negative work (or zero work) required to achieve the starting point.
This implies that the line integral of $\overrightarrow{F}$ is not path independent and thus, the vector field $\overrightarrow{F}$ is not conservative.