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I've been having a lot of trouble understanding the statement of the theorem and its proof, so I would like to see if I did the following question below correctly.

**The problem**

Let f : R² → R be given by f(x,y,z) = sin(xyz) + e^[2x + y(z - 1)]. Show that the level set {f = 1} can be solved as x = x(y,z) near (0,0,0) and compute ∂x/∂y (0,0) and ∂x/∂z (0,0).

**SOLUTION!**

Let G(x,y,z) = f(x,y,z) - 1 and note that G(0,0,0) = 0.

DG = [yzcos(xyz) + e^[2x + y(z - 1)]; xzcos(xyz) + (z - 1)e^[2x + y(z - 1)]; xycos(xyz) + e^[2x + y(z - 1)]; 1 x 3 matrix

I have ∂G/∂x = yzcos(xyz) + 2e^[2x + y(z - 1)]. Now at (0,0,0) I have it equal to 1 and hence the determinant is nonzero, so I can apply the Implicit Function Theorem.

So Dx(0,0) = -[∂G/∂x]^(-1) ⋅ [∂G/∂(y,z)] = -[1]^(-1) ⋅ [-1 1] = [1 -1]

Therefore, ∂x/∂y (0,0) = 1 and ∂x/∂z (0,0) = -1.

Is this all right?