Let $M$ be set of points $(x,y,z) \in \mathbb{R}^3$ such that $\lvert z-x \rvert = \sqrt{x+y^2}$. Is $M$ two dimensional manifold?
I have two different solutions to this problem and in one of them the result is that $\lvert z-x \rvert = \sqrt{x+y^2}$ is manifold but in the second I got that it is not manifold. Clearly one of solution is wrong, I do not know which and I do not know why.
First solution.
Raise both sides of equation to second power and consider function $$ F(x,y,z) = z^2 - 2xz + x^2 - x - y^2 $$
Gradient of $F$ is: $$ DF(x,y,z) = (-2z+2x-1,\, -2y,\, 2z-2x), $$ which is nonzero for all $(x.y,z)$ so $DF(x,y,z)$ is epimorphism in every $(x,y,z)$ and therefore $M$ is two dimensional manifold. (This is by theorem I learned in my analysis class.)
Second solution.
Fix $x=0$, we get $\lvert z \rvert = \lvert y \rvert$ so none of $y$ and $z$ are functions of two other variables.
Fix $z=0$, we get $x^2-x = y^2$ and when I graph it on Desmos, I see that for many $x$'s there are two $y$'s so $x$ is not a function of $y$ and $z$.
So there are points in $M$ where $M$ is not a graph of a function so $M$ is not manifold.
The first solution is correct.
The last sentence of the second solution is not. You cannot "fix $x=0$" for instance. You can only choose some $(0,y,z)∈M,$ but in a neighborhood in $M$ of that point, $x$ is not fixed and (by the first solution) $M$ is a graph of a function.
E.g. in a neighborhood in $M$ of $(0,0,0),$ $$x=\frac12\left(1+2z-\sqrt{1+4z+4y^2}\right).$$