Slovenian Mathematical Olympiad, 2016

Denote $a = \log_3 x^2$ and $b = \log_2 y^3$. Then
$$\log_9 x^4 = \frac{\log_3 x^4}{\log_3 9} = \frac{2 \log_3 x^2}{2} = a \quad \text{and} \quad \log_4 y^9 = \frac{\log_2 y^9}{\log_2 4} = \frac{3 \log_2 y^3}{2} = \frac{3}{2}b.$$
Inserting this into the initial equations we get $a+b=1$ and $a+\frac{3}{2}b=2$. Subtracting the first equation from the second yields $\frac{1}{2}b=1$ or $b=2$. From the first equation we get $a=-1$. From $\log_3 x^2 = -1$ it follows that $x^2 = \frac{1}{3}$ or $x = \pm\frac{1}{\sqrt{3}}$, and from $\log_2 y^3 = 2$ we have $y^3 = 4$ or $y = \sqrt[3]{4}$.