level surface - question from mathematics analytics 2 Can someone help me with this exercise? I don't know how to solve it.
Consider the function $f:Df ⊂ \mathbb{R}^3\rightarrow \mathbb{R}$, $f(x,y,z) = \frac{1}{\sqrt[3]{x^3-y}}-e^{z-x}$. 
The graphic of the function  $g: Dg ⊂ \mathbb{R^3}\rightarrow \mathbb{R}^3$, $g(x,y) = x - \ln(\sqrt{x^3-y})$ is a level surface  $c$ of the function  $f$, for which value from $c$?
 A: The following is merely pointing out that Ovi and Jean Marie have together effectively answered the question already. Unfortunately it's too long for a comment.
The reason for Ovi's question (presumably) is that if you're taking $\ g\ $ to be a function of the three variables $\ x, y, z\ $, then its graph is, by definition, the set $\ \{(x,y,z, g(x,y))\,\vert\, (x,y,z)\in \mathbb R^3\,\}$, a 3-dimensional surface in the 4-dimensional space $\ \mathbb R^4\ $.  This cannot be a level surface of $\ f\ $, which has to be a 2-dimensional surface of the form $\ \{ (x,y,z)\,\vert\, f(x,y,z) = k\ \}$, lying in the 3-dimensional space $\ \mathbb R^3\ $.
If, on the other hand, you were to take $\ g\ $ to be a function of just the two variables $\ x,y\ $, then what you'd have to show would be that the set $\ \{(x,y, g(x,y))\,\vert\, (x,y)\in \mathbb R^2\,\}$ (i.e. the graph of $\ g\ $) is identical to $\ \{ (x,y,z)\,\vert\, f(x,y,z) = k\ \}$ for some value of $\ k\ $, which Jean Marie's comment is suggesting to you might be true for the value $\ k = 0\ $.
