There isn't enough information, clearly

$\begin{aligned} x^2 - 3y^2 &= 2xy\\ x^2 - y^2 &= 2xy + 2y^2\\ (x-y)(x+y) &= 2y(x+y)\\ x-y &= 2y\\ x &= 3y\\ \implies \dfrac{x}{y} &= 3\\ \end{aligned}$

$\begin{aligned} x^2-3y^2 & = 2xy \\ x^2-2xy - 3y^2 & = 0 \\ x^2-2xy + y^2 & = 4y^2 \\ (x-y)^2 & = (2y)^2 \\ x-y & = 2y \\ x & = 3y \\ \implies \frac xy & = \boxed{3} \end{aligned}$

Divide both sides by xy, then we have (x/y) - 3(y/x) = 2, or (x/y)^2 -2(x/y)-3=0. We can factorize to ((x/y)-3)((x/y+1) = 0. So the solutions are x/y = 3, or -1. Since x, and y are both positive than x/y=3

$x^2-3y^2=2xy$

$x^2-3y^2-2xy=0$

$(x+y)(x-3y)=0$

$x-3y=0$

$x=3y$

$\Rightarrow\dfrac{x}{y}=\boxed3$

A handy trick goes as follows: since y>0, divide the equation by y² to get: (x/y)² -2(x/y) -3 = 0 which is a quadratic equation in x/y with 3 as the only positive solution.