Level surface undefined Can a level surface be undefined at some point, even if the original fuction is defined at the same point?
example:
$w(x,y,z) = xy+yz+xz$ is defined at $p=(1,-1,2).$
Its level surface at $p$ is  $z=-\frac{1}{(x+y)}-\frac{xy}{(x+y)},$ which is NOT defined at $(1,-1,2),$ since $x+y = 0.$
How can a function be defined at some point and NOT its level surface?
 A: As $ w(1,-1,2) = -1 $, I suppose that you want to study the set
$$
S = \left\{ (x,y,z) \in \mathbb{R}^{3} ~ \middle| ~ x y + (x + y) z = -1 \right\}.
$$
We call $ S $ the level-set of $ w $ corresponding to $ -1 $.
For $ (x,y,z) \in S $, there are two cases:


*

*$ x + y \neq 0 $, in which case $ z = - \dfrac{1 + x y}{x + y} $.

*$ x + y = 0 $, in which case $ x y = -1 $ and $ z $ can take any real value.


In the second case, the conditions $ x + y = 0 $ and $ x y = -1 $ yield
$$
(x,y) = (-1,1) \quad \text{or} \quad (x,y) = (1,-1).
$$
We therefore have
\begin{align}
    S
= ~ & \left\{
      (x,y,z) \in \mathbb{R}^{3} ~ \middle| ~
      x + y \neq 0 ~ \text{and} ~ z = - \frac{1 + x y}{x + y}
      \right\} \cup \\
    & \left\{ (-1,1,z) \in \mathbb{R}^{3} ~ \middle| ~ z \in \mathbb{R} \right\}
      \cup \\
    & \left\{ (1,-1,z) \in \mathbb{R}^{3} ~ \middle| ~ z \in \mathbb{R} \right\}.
\end{align}
The last two sets account for the points $ (x,y,z) \in S $ where $ x + y = 0 $.
Hence, the answer to your question is: Any point in $ \mathbb{R}^{3} $ belongs to the level-set corresponding to the value of $ w $ at that point. However, it is not possible to write a level set of $ w $ in the form
$$
z = F(x,y).
$$
