az/ax=az/auau/ax+az/avav/ax=az/au+az/av(-y/x^2)
az/ay=az/auau/ay+az/avav/ay=az/av(1/x)
xaz/ax+yaz/ay=xaz/au+az/av(-y/x)+az/av(y/x)=xaz/au=uaz/au
xaz/ax+yaz/ay=z
即z=uaz/au
U'(x)=U'(m)m'(x)+U'(n)n'(x)=U'(m)+U'(n)；
U'(y)=U'(m)m'(y)+U'(n)n'(y)=aU'(m)+bU'(n)；
所以：U''(xx)=U''(mm)+2U''(mn)+U''(nn)；U''(xy)=aU''(mm)+bU''(mn)+aU''(mn)+bU''(nn)
U''(yy)=aU''(mm)a+aU''(mn)b+bU''(mn)a+bU''(nn)b
0=6Uxx-5Uxy+Uyy=6(U''(mm)+2U''(mn)+U''(nn))-5(aU''(mm)+bU''(mn)+aU''(mn)+bU''(nn))+
aU''(mm)a+aU''(mn)b+bU''(mn)a+bU''(nn)b
=(6-5a+a^2)U''(mm)+(12-5b-5a+2ab)U''(mn)+(6-5b+b^2)U''(nn)
由题意：6-5a+a^2=0,6-5b+b^2=0,解得：a=2,b=3或a=3,b=2或a=b=2或a=b=3
u=x,v=y/x
那么得到az/ax=az/auau/ax+az/avav/ax
=az/au+az/av(-y/x²)
而az/ay=az/auau/ay+az/avav/ay
=az/av1/x
代入xaz/ax+yaz/ay=z中
得到z=uaz/au+az/av(-y/x)+az/avy/x=uaz/au
就得到了证明
au/ax=au/aξaξ/ax+au/aηaη/ax+au/aζaζ/ax=au/aξ-au/aη-au/aζ
au/ay=au/aξaξ/ay+au/aηaη/ay+au/aζaζ/ay=au/aη
au/az=au/aξaξ/az+au/aηaη/az+au/aζaζ/az=au/aζ
所以au/ax+au/ay+au/az=au/aξ-au/aη-au/aζ+au/aη+au/ζ=au/aξ=0
命题得证
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