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?
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:
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 $$ z = F(x,y). $$
