Estonia — Open Contests

By symmetry we can assume that $x > y$. If $x - y = 1$, then
$$\begin{aligned} \frac{x^2 + 4xy + y^2}{x^3 - y^3} &= \frac{x^2 + 4xy + y^2}{(x-y)(x^2 + xy + y^2)} = \frac{(x-y)^2 + 6xy}{(x-y)((x-y)^2 + 3xy)} = \\ &= \frac{1 + 6xy}{1 + 3xy} = 1 + \frac{3xy}{1 + 3xy}, \end{aligned}$$
which is clearly not an integer. If $x - y \ge 2$, then
$$\begin{aligned} \frac{x^2 + 4xy + y^2}{x^3 - y^3} &= \frac{x^2 + 4xy + y^2}{(x-y)(x^2 + xy + y^2)} \le \frac{x^2 + 4xy + y^2}{2(x^2 + xy + y^2)} < \\ &< \frac{2x^2 + 2xy + 2y^2}{2(x^2 + xy + y^2)} = 1, \end{aligned}$$
where the last inequality follows from $x^2 - 2xy + y^2 = (x-y)^2 > 0$.

If $\frac{x^2 + 4xy + y^2}{x^3 - y^3}$ were an integer, then
$$\frac{x^2 + 4xy + y^2}{x^3 - y^3} \cdot (x - y) - 1 = \frac{3xy}{x^2 + xy + y^2}$$

$$\begin{aligned} \frac{x^2 + 4xy + y^2}{x^3 - y^3} &= \frac{x^2 + 4xy + y^2}{(x-y)(x^2 + xy + y^2)} \le \frac{x^2 + 4xy + y^2}{2(x^2 + xy + y^2)} < \\ &< \frac{2x^2 + 2xy + 2y^2}{2(x^2 + xy + y^2)} = 1, \end{aligned}$$
$$$$
where the last inequality follows from $x^2 - 2xy + y^2 = (x-y)^2 > 0$.