Hinz? =? xyz？ g(x，y，z)=sinz-x？ yz = 0； Meet the following three conditions:

g'(x)=2xyz，g'(y)=-x？ z，g'(z)=cosz-x？ Y is continuous in the neighborhood of (x0, y0, z0); Subject: (x0y0z0)=(000)

g(x0，y0，z0)=0

g'(z)(x0，y0，z0)= 1≠0

Then a unique single-valued function z=f(x, y) exists in the neighborhood of (x0, y0, z0) and has the following properties:

g[x，y，f(x，y)]=0，？ f(x0，y0)=z0

F(x, y) continuity

F(x, y) has continuous partial derivatives:

z？ “x=-g，”x/g？ z； “z，”y=-g？ “y/g，”z

This is the existence theorem of multivariate implicit function, and the proof is complicated. You can refer to related books.

Let's find the partial derivative:

z'x=-g'x/g'z=-2xyz/(cosz-x？ y) z'x(0，0，0)= 0；

z'y=-g'y/g'z=-x？ z/(cosz-x？ y)？ z'y(0，0，0)=0。