Slovenian Mathematical Olympiad, 2008

The second equation implies $y = x^{-2}$, so
$$x^{x + x^{-2}} = x^{-2(x - x^{-2})}.$$
Taking the logarithm on both sides we get
$$(x + x^{-2}) \log x = -2(x - x^{-2}) \log x.$$
If $\log x = 0$, then $x = 1$ and $y = 1$. Otherwise,
$$x + x^{-2} = -2x + 2x^{-2},$$
so $3x^3 = 1$. This implies $x = \frac{1}{\sqrt{3}}$ and $y = \sqrt{3}$. There are two solutions: $x = 1$, $y = 1$ and $x = \frac{1}{\sqrt{3}}$, $y = \sqrt{3}$.