Obviously, $|x| > |y|$, for $x \neq 0$, since otherwise, the first fraction will not be an integer. If the first two expressions are equal, then we have the equality:
$$\frac{x}{y} = \frac{x^2+x}{y^2+y} \Rightarrow xy^2 + xy = yx^2 + xy \Rightarrow xy(y-x) = 0.$$
For $x \neq y$ and possible range of values, the given equality is satisfied only by pairs $(x, y) = (0, t)$, $t \neq 0; -1$. The third expression will obviously take a non-zero, but non-integer value, since $0 < \frac{2}{t^2+2} < 1$ for $t \neq 0$.

If the first and third expressions are equal, we have the equality:
$$\frac{x}{y} = \frac{x^2+2}{y^2+2} \Rightarrow xy^2 + 2x = yx^2 + 2y \Rightarrow xy(y-x) = 2(y-x) \Rightarrow xy = 2,$$
The possible pairs of $(x, y)$ are (2, 1) and (-2, -1). Here, the second expression in the first case will take on a value 3, hence, different from the other two. In the second case, it is not defined and, therefore, not possible.

If the second and third expressions are equal, we have the equality:
$$\frac{x^2+x}{y^2+y} = \frac{x^2+2}{y^2+2} \Rightarrow x^2y^2 + xy^2 + 2x^2 + 2x = x^2y^2 + yx^2 + 2y^2 + 2y \Rightarrow \\ xy(y-x) = 2(y-x)(y+x) + 2(y-x) \Rightarrow xy - 2y - 2x = 2 \Rightarrow (x-2)(y-2) = 6.$$

$x-2=6, y-2=1 \Rightarrow x=8, y=3 \Rightarrow \frac{x^2+2}{y^2+2} = \frac{66}{11} = 6$ and $\frac{x}{y} = \frac{8}{3}$ does not satisfy the conditions.

$x-2=3, y-2=2 \Rightarrow x=5, y=4 \Rightarrow \frac{x^2+2}{y^2+2} = \frac{27}{18}$ is not integer, so does not satisfy the conditions.
$x-2=-2, y-2=-3 \Rightarrow x=0, y=-1$ does not lie in the allowed range of values.
$x-2=-6, y-2=-1 \Rightarrow x=-4, y=1 \Rightarrow \frac{x^2+2}{y^2+2} = \frac{18}{3} = 6$ and $\frac{x}{y} = -4$ satisfies the conditions.