Answer
Not conservative.
Work Step by Step
When $F(x,y)=Ai+Bj$ is a conservative field, then throughout the domain $D$, we get
$\dfrac{\partial A}{\partial y}=\dfrac{\partial B}{\partial x}$
$a$ and $b$ are the first-order partial derivatives on the domain $D$.
Since, $\int_C \overrightarrow{F} \cdot \overrightarrow{dr}$ is independent of path if and only if $\int_C \overrightarrow{F} \cdot \overrightarrow{dr}=0$ for every closed curve $C$.
The work done is a line integral of force. When a closed loop is drawn around the center of the vector field, there will be a non-zero amount of work needed to get to the initial point.
However, in a conservative field, there will be an equal amount of positive and negative work (or zero work) needed to get to the initial point.
This means that the line integral of $\overrightarrow{F}$ is not path independent.
Hence, the vector field $\overrightarrow{F}$ is not conservative.
