Answer
$\mathbf{F}$ is conservative
Work Step by Step
Given $$\mathbf{F}(x, y)=\frac{2}{y^2} e^{\frac{2x}{y}}(y \mathbf{i}- x \mathbf{j})=\frac{2}{y} e^{\frac{2x}{y}}\mathbf{i}-\frac{2x}{y^2} e^{\frac{2x}{y}}\mathbf{j} $$ 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{2}{y} e^{\frac{2x}{y}} } \\ {N=f_{y}(x, y)=\frac{\partial f(x,y)}{\partial y}=-\frac{2x}{y^2} e^{\frac{2x}{y}} } \\ \end{array} this implies that $M$ and $N$ have continuous first partial derivatives at $ y\ne0$. Since \begin{array}{l} { \frac{\partial M}{\partial y}=-\frac{2}{y^2} e^{\frac{2x}{y}}+\frac{2}{y} \frac{-2x}{y^2} e^{\frac{2x}{y}}=-\frac{2}{y^3} (y+2x)e^{\frac{2x}{y}} } \\ { \frac{\partial N}{\partial x}=-\frac{2}{y^2} e^{\frac{2x}{y}}+ \frac{-2x}{y^2}\frac{2}{y} e^{\frac{2x}{y}}=-\frac{2}{y^3} (y+2x)e^{\frac{2x}{y}}} \\ \end{array} So, we get $$ \frac{\partial N}{\partial x}= \frac{\partial M}{\partial y}$$ and this implies that $\mathbf{F}$ is conservative