Don't quite understand question

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In summary, the level set \{f=1\} can be solved as x=x(y,z) near (0,0,0) using the implicit function theorem. We can define the function X(y,z) to be 1-y-z, which satisfies the condition that F(y,z)=f(X(y,z),y,z)=1 for all (y,z) in a region D around (0,0,0). This can be seen by taking an example of f(x,y,z)=x+y+z and setting f=1, which results in X(y,z)=1-y-z and F(y,z)=1. This can be applied to find the partial derivatives \frac{\partial x}{\partial y} (0
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JG89
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Question: Let [tex] f: \mathbb{R}^3 \rightarrow \mathbb{R} [/tex] be given by [tex] f(x,y,z) = sin(xyz) + e^{2x + y(z-1)} [/tex]. Show that the level set [tex] \{ f = 1 \} [/tex] can be solved as [tex] x = x(y,z) [/tex] near [tex] (0,0,0) [/tex] and compute [tex] \frac{\partial x}{\partial y} (0,0) [/tex] and [tex] \frac{\partial x}{\partial z} (0,0,0) [/tex].

I don't understand this question. I know what the level set is. It will be the set of points (x,y,z) such that f(x,y,z) = 1. I don't understand what they mean by, show that the level set can be solved as x = x(y,z) near (0,0,0). Any help?
 
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Now, if f(x,y,z)=1, the implicit function theorem says that in some region D in the (y,z)-domain around a solution vector [tex](x_{0},y_{0},z_{0})[/tex], we may define a function X(y,z), so that within D, the function F(y,z)=f(X(y,z),y,z)=1 for ALL (y,z) in D.

To take an example.

1. Let f(x,y,z)=x+y+z

Set f=1.

We now define: X(y,z)=1-y-z

Note that we then have:
F(y,z)=f(X(y,z),y,z)=(1-y-z)+y+z=1, irrespective of the values of (y,z).

This holds, for example, in some neighbourhood of the point (1,0,0).
 

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