If #x+y+z=0#,
#z=-x-y#, so #yz=-xy-y^2#.
With same logic, #xz=-yz-z^2# and #xy=-xz-x^2#
Hence,
#(yz)/(2x^2+yz)+(xz)/(2y^2+xz)+(xy)/(2z^2+xy)#
=#(yz)/(2x^2-xy-y^2)+(xz)/(2y^2-yz-z^2)+(xy)/(2z^2-xz-x^2)#
=#(yz)/[(2x+y)(x-y)]+(xz)/[(2y+z)(y-z)]+(xy)/[(2z+x)(z-x)]#
=#(yz)/[(x+x+y+z-z)(x-y)]+(xz)/[(y+y+z+x-x)(y-z)]+(xy)/[(z+z+x+y-y)(z-x)]#
=#(yz)/[(x-z)(x-y)]+(xz)/[(y-x)(y-z)]+(xy)/[(z-y)(z-x)]#
=#(yz)/[(x-z)(x-y)]-(xz)/[(x-y)(y-z)]+(xy)/[(y-z)(x-z)]#
=#[yz*(y-z)-xz*(x-z)]/[(x-y)(y-z)(x-z)]+(xy)/[(y-z)(x-z)]#
=#(y^2z-yz^2-x^2z+xz^2)/[(x-y)(y-z)(x-z)]+(xy)/[(y-z)(x-z)]#
=#[(x-y)z^2-z(x^2-y^2)]/[(x-y)(y-z)(x-z)]+(xy)/[(y-z)(x-z)]#
=#[(x-y)z^2-z(x+y)(x-y)]/[(x-y)(y-z)(x-z)]+(xy)/[(y-z)(x-z)]#
=#[(x-y)z(z-x-y)]/[(x-y)(y-z)(x-z)]+(xy)/[(y-z)(x-z)]#
=#[z*(z+z-x-y-z)]/[(y-z)(x-z)]+(xy)/[(y-z)(x-z)]#
=#(z*2z)/[(y-z)(x-z)]+(xy)/[(y-z)(x-z)]#
=#(2z^2+xy)/[(y-z)(x-z)]#
=#(2z^2-xz-x^2)/[(y-z)(x-z)]#
=#[(2z+x)(z-x)]/[(y-z)(x-z)]#
=#[(z+z+x+y-y)(z-x)]/[(y-z)(x-z)]#
=#[(z-y)(z-x)]/[(y-z)(x-z)]#
=#[(y-z)(x-z)]/[(y-z)(x-z)]#
=#1#
