Answer
$\mathbf{F}$ is conservative
Work Step by Step
Given
$$\mathbf{F}(x, y)=\frac{1}{xy}(y \mathbf{i}- x \mathbf{j})=\frac{1}{x}\mathbf{i}-\frac{1}{y} \mathbf{j} $$
Since \begin{align}\mathbf{F}(x,y)&=M \mathbf{i}+N\mathbf{j}\\
&=f_{x}(x, y)\mathbf{i}+f_{y}(x, y)\mathbf{j}
\end{align}
We get
\begin{array}{l}
{M=f_{x}(x, y)=\frac{\partial f(x,y)}{\partial x}=\frac{1}{x} } \\
{N=f_{y}(x, y)=\frac{\partial f(x,y)}{\partial y}=-\frac{1}{y} } \\ \end{array}
This implies that $M$ and $N$ have continuous first partial
derivatives at $x\ne0, y\ne0$.
Since
\begin{array}{l} { \frac{\partial M}{\partial y}=0}
\\ { \frac{\partial N}{\partial x}=0} \\ \end{array}
So, we get
$$ \frac{\partial N}{\partial x}= \frac{\partial M}{\partial y}$$
And this implies that $\mathbf{F}$ is conservative