1/x+1/y+1/z=0 xy/z^2+yz/x^2+zx/y^2=3 доказать. помогите

Решите задачу:

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z}=0\quad \to \quad \frac{yz+zx+xy}{xyz} =0 \quad \to\; \; xy+yz+zx=0\\\\\\ Dokazat:\\\\\frac{xy}{z^2} +\frac{yz}{x^2} + \frac{zx}{y^2} =3\; .$

$Formyla:\\\\a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc\\\\(xy)^3+(yz)^3+(zx)^3=\underbrace {(xy+yz+zx)}_{0}(x^2y^2+y^2z^2+z^2x^2-xy^2z-\\\\-yz^2x-zx^2y)+3x^2y^2z^2$

$(xy)^3+(yz)^3+(zx)^3=3x^2y^2z^2\; |:x^2y^2z^2\ne 0\\\\ \frac{(xy)^3}{x^2y^2z^2} + \frac{(yz)^3}{x^2y^2z^2} + \frac{(zx)^3}{x^2y^2z^2} =3\\\\ \frac{xy}{z^2} + \frac{yz}{x^2} + \frac{zx}{y^2}=3$