Solution:
$$\frac{-1 + \sqrt{7}}{2}$$
We have $3 = k^{2}\left(\frac{x^{2}}{y^{2}} + \frac{y^{2}}{x^{2}}\right) + k\left(\frac{x}{y} + \frac{y}{x}\right) \geq 2k^{2} + 2k$, hence $7 \geq 4k^{2} + 4k + 1 = (2k + 1)^{2}$, hence $k \leq (\sqrt{7} - 1)/2$. Obviously $k$ can assume this value, if we let $x = y = 1$.