Belarus — Selection and Training Session

First, note that $a \neq 0$, $x \neq -y$, etc. (this easily follows from the given relation). Now, the equality $x^2 - 1/y = y^2 - 1/z$ can be written in two other ways:
$$\begin{aligned} x^2 - y^2 + \frac{1}{x} - \frac{1}{y} &= \frac{1}{x} - \frac{1}{z} \quad \Leftrightarrow \quad (x-y)(x+y-\frac{1}{xy}) = \frac{z-x}{xz} \\ (x-y)(x^2+xy-\frac{1}{y}) &= \frac{z-x}{z} \quad \Leftrightarrow \quad (x-y)(a+xy) = \frac{z-x}{z} \end{aligned} \quad (1)$$
and
$$\begin{aligned} x^2 - y^2 - \frac{1}{x} - \frac{1}{y} &= \frac{1}{x} - \frac{1}{z} \quad \Leftrightarrow \quad (x+y)(x-y-\frac{1}{xy}) = -\frac{z+x}{xz} \\ (x+y)(x^2 - xy - \frac{1}{y}) &= -\frac{z+x}{z}. \end{aligned} \quad (2)$$
Now, multiplying (1) by two similar equalities and reducing by $(x-y)(y-z)(z-x) \neq 0$, we obtain $(a+xy)(a+yz)(a+zx) = \frac{1}{xyz}$
$$a^3 + a^2(xy + yz + zx) + a(x + y + z)xyz + (xyz)^2 = \frac{1}{xyz}. \quad (3)$$
Proceeding similarly with (2), we obtain
$$a^3 - a^2(xy + yz + zx) + a(x + y + z)xyz - (xyz)^2 = -\frac{1}{xyz}. \quad (4)$$
Summing (3) and (4) and reducing by $2a \neq 0$, we obtain the required equality.