55rd Ukrainian National Mathematical Olympiad - Third Round

Denote $a=x^2-y^2$, $b=y^2-z^2$, then $-a-b=z^2-x^2$ and the equation can be rewritten as: $$\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}\iff (a+b{)}^2=ab\iff a^2-ab+b^2=0\iff {\left(a-\frac{b}{2}\right)}^2+\frac{3b^2}{4}=0.$$ The last equality holds only with $a=b=0$, which is impossible. Therefore, the equation has no solutions.