Let $f:\mathbb{R}^3\to\mathbb{R}, f(x,y,z)=(x-y+z-1)^2$. For which values of $a$ is $\{(x,y,z)\in\mathbb{R}^3:f(x,y,z)=a\}$ a 2-manifold?
Instead of $(x-y+z-1)^2=a$ seems a better idea to write $x-y+z-1=\sqrt{a}$ so we can have $x=y-z+\sqrt{a}+1$ and the question becomes: for which values of $a$ is $\{(y-z+\sqrt{a}+1,y,z):y,z\in\mathbb{R}\}$ a 2-manifold?.
But how to solve this?. I know I need to find a diffeomoprhism from the set to $\mathbb{R}^2$, but I'm having troubles due to the generality of the problem, this is, how could I show that is not possible to define a diffeomorphism from the set to $\mathbb{R}^2$ for any value of $a$?.
Maybe if I put some values I can get a few solutions, but seems a lot harder to show that it is not possible to define any function to be a diffeomorphism than to find it for certain values of $a$.
You can apply the change of coordinates $(x,y,z)\mapsto (x,x-y+z-1,z)$ (which is involutive), and your question concerns $\{(x,y,z)\in \mathbb{R}^3\mid y^2=a\}$.
If $a$ is positive, you get two planes. If $a$ is zero, you get one plane and if $a$ is negative, then you get the empty set.