Answer
$\mathbf{F}$ is not conservative.
Work Step by Step
Given
$$\mathbf{F}(x, y)=\frac{2 y}{x} \mathbf{i}-\frac{x^{2}}{y^{2}} \mathbf{j}$$
Since \begin{align}\mathbf{F}(x,y)&=M \mathbf{i}+N\mathbf{j}\end{align}
We get
\begin{array}{l}
{M=\frac{2 y}{x} } \\
{N= - \frac{ x^2}{y^2} } \end{array}
This implies that $M$ and $N$ have continuous first partial
derivatives.
As, we have
\begin{array}{l} {\begin{align} \frac{\partial M}{\partial y}&=\frac{2 }{x}
\end{align}}
\\ {\begin{aligned} \frac{\partial N}{\partial x}&=-\frac{2x}{y^2} \end{aligned}} \\
\end{array}
So, we get
$$ \frac{\partial N}{\partial x}\neq \frac{\partial M}{\partial y}$$
And this implies that $\mathbf{F}$ is not conservative