1 q? x/？ Y: if x=x(y, z) is substituted into the equation F(x, y, z)=0, that is, F(x(y, z), y, z)=0, it is regarded as an equation about the unknown number y, z, and then the derivatives of its two sides about y, F 1'*? x/？ Y+F2'* 1+F3'*0=0, so? x/？ y =-F2 '/f 1 '；

2 q? y/？ Z: then substitute y=y(x, z) into the equation F(x, y, z)=0, that is, F(x, y(x, z), z)=0, and then treat it as an equation about the unknown number x, z, and take the derivative of both sides of z to get f 1' * 0+F2. y/？ Z+F3'* 1=0, so? y/？ z =-F3 '/F2 '；

3 q? z/？ X: if z=z(x, y) is still substituted into the equation F(x, y, z)=0, that is, F(x, y, z(x, y))=0, it is regarded as an equation about the unknown number x, y, and its two sides are differentiated with respect to x, and F 1'* 65438 is obtained. z/？ X=0, so? z/？ x =-f 1 '/F3 '；

So (? x/？ y)*(？ y/？ z)*(？ z/？ x)=(-F2 '/f 1 ')(-F3 '/F2 ')(-f 1 '/F3 ')=- 1。

To sum up, (? x/？ y)*(？ y/？ z)*(？ z/？ x)=- 1 .