Ukrajina, 2008

Let $B = \frac{1}{x-y} + \frac{1}{y-z} + \frac{1}{z-x} = \frac{3}{2}$ and $A = \frac{1}{(x-y)^2} + \frac{1}{(y-z)^2} + \frac{1}{(z-x)^2}$ then
$$B^2 = \frac{9}{4} = A + 2\left(\frac{1}{(x-y)(y-z)} + \frac{1}{(x-y)(z-x)} + \frac{1}{(y-z)(z-x)}\right).$$
Let's rearrange the second summand of the last expression:
$$\frac{1}{(x-y)(y-z)} + \frac{1}{(x-y)(z-x)} + \frac{1}{(y-z)(z-x)} = \frac{(z-x)+(y-z)+(x-y)}{(x-y)(y-z)(z-x)} = 0.$$
Therefore, $A = \frac{9}{4}$.

Though it is not required in the problem, we should check that there are numbers $x$, $y$, $z$ which satisfy the given condition. Let $x=2$, $y=1$, $z=0$ then $(x - y) = 1$, $(y - z) = 1$, $(z - x) = -2$, $B = 1 + 1 - \frac{1}{2} = \frac{3}{2}$, and $A = 1 + 1 + \frac{1}{4} = \frac{9}{4}$.