National Math Olympiad - Final Round, 2013

Taking logarithms on both sides of the equation and taking into account that $\log a^b = b \log a$ and $\log_a b = \frac{\log b}{\log a}$, we get
$$\frac{\log x \log 2013}{\log 3} = \frac{\log y \log 2013}{\log 5},$$
or
$$\log y = \frac{\log 5}{\log 3} \log x = \log_3 5 \log x.$$
From here we can conclude that $\log y$ and $\log x$ have the same sign, so $x$ and $y$ are either both less than $1$, both equal to $1$ or both greater than $1$.

On the other hand the inequality can be rewritten as $\log_{\frac{1}{2}}(xy) > 0$ and since $\frac{1}{2} < 1$ we have $xy < 1$.

Both together imply that $x$ and $y$ are less than $1$, so $\log x$ and $\log y$ are negative. But $\log_3 5 > 1$ implies that $\log y = \log_3 5 \log x < \log x$, or $y < x$.

So, $x$ is greater.