Answer
$\mathbf{F}$ is conservative
Work Step by Step
Given
$$\mathbf{F}(x, y)= \frac{1}{x^{2}}(y \ \mathbf{ i}-x \ \mathbf{j})=\frac{y}{x^{2}}\ \mathbf{ i}-\frac{1}{x} \ \mathbf{j} $$
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{y}{x^{2}}} \\ {N=f_{y}(x, y)=\frac{\partial f(x,y)}{\partial y}=-\frac{1}{x}} \\ \end{array}
This implies that $M$ and $N$ have continuous first partial
derivatives at all $x\ne 0$.
Since
\begin{array}{l} { \frac{\partial M}{\partial y}=\frac{1}{x^{2}}}
\\ { \frac{\partial N}{\partial x}=\frac{1}{x^{2}}} \\ \end{array}
So, we get
$$ \frac{\partial N}{\partial x}= \frac{\partial M}{\partial y}$$
And this implies that $\mathbf{F}$ is conservative