Slovenija 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 $x=\frac{1}{\sqrt{3}}$, $y=\sqrt{3}$.