Ukrainian National Mathematical Olympiad

Obviously, $x \neq y$, because otherwise, all three expressions would take the same value, which contradicts problem statement.
If the first two expressions are equal, then we have the equality:
$$\frac{x}{y} = \frac{x^2+4}{y^2+4} \Rightarrow xy^2 + 4x = yx^2 + 4y \Rightarrow xy(y-x) = 4(y-x).$$
Since $x \neq y$, we can divide by $y-x \Rightarrow xy=4$. There are two pairs of positive integer numbers that satisfy the last equality: (1, 4) and (4, 1). Substitute them into the third equality, and we get
$$\frac{x^3+8}{y^3+8} = \frac{9}{72} = \frac{1}{8} \neq \frac{x}{y} \quad \text{and, analogously,} \quad \frac{x^3+8}{y^3+8} = 8 \neq \frac{x}{y}, \quad \text{so these pairs of numbers satisfy the}$$
condition.

If the first and third expressions are equal, then
$$\frac{x}{y} = \frac{x^3 + 8}{y^3 + 8} \Rightarrow xy^3 + 8x = yx^3 + 8y \Rightarrow xy(y-x)(y+x) = 8(y-x) \Rightarrow xy(y+x) = 8,$$
which has no solution, as one can confirm by a simple search.
If the second and third expressions are equal, then we have the equality:
$$\begin{aligned} \frac{x^2+4}{y^2+4} &= \frac{x^3+8}{y^3+8} \Rightarrow x^2y^3 + 8x^2 + 4y^3 = x^3y^2 + 8y^2 + 4x^3 \\ x^2y^2(y-x) + 4(y-x)(y^2 + xy + x^2) &= 8(y-x)(y+x) \Rightarrow x^2y^2 + 4(y^2 + xy + x^2) = 8(y+x) \Rightarrow 4xy + x^2y^2 = 4(2y-y^2) + 4(2x-x^2). \end{aligned}$$

So, $x=1$ or $y=1$. If $x=1$, then $4y+y^2 = 4(2y-y^2)+4$ or $5y^2-3y+4=0$ -
equality is obviously not possible, because $5y^2 > 3y$ for positive integer $y$. Analogously for $y=1$.