Answer
The vector field is 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 $D$.
$\int_C \overrightarrow{F} \cdot \overrightarrow{dr}$ 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. In the given graph it has been seen that when a closed loop is drawn in the direction 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 sum of these vectors will have the resultant vector having magnitude of $0$. This can only be possible when the vector field is conservative.
Hence, the vector field is conservative.