2015/103) If $xy + yz + zx =0$ then $\dfrac{1}{x^2 - yz} + \dfrac{1}{y^2 - zx} +\dfrac{1}{z^2 - xy}$ is equal to

$xy + yz + zx = 0$

so $yz = -x(y+z)$
$x^2-yz = x(x+y+z)$

so $\dfrac{1}{x^2-yz} = \dfrac{1}{x(x+y+z)}$
similarly$\dfrac{1}{y^2-xz} = \dfrac{1}{y(x+y+z)}$
and
$\dfrac{1}{z^2-xy} = \dfrac{1}{z(x+y+z)}$

adding all 3 we get your expression

$\dfrac{1}{x^2 - yz} + \dfrac{1}{y^2 - zx} +1\dfrac{1}{z^2 - xy} = \dfrac{1}{x+y+z}(\dfrac{1}{x}+\dfrac{1}{y} + \dfrac{1}{z})$

$\dfrac{1}{x} +\dfrac{1}{y} + \dfrac{1}{z} = \dfrac{yz+xz+xy}{xyz} = 0$

so
$\dfrac{1}{x^2 - yz} + \dfrac{1}{y^2 - zx} +\dfrac{1}{z^2 - xy} = 0$