I've seen an exercise in which I am to prove that $$\{(x,y,z) \in \mathbb{R}^3 : x^2+y^2=z^2\}$$ is not a submanifold of $\mathbb{R}^3$. I've done a little bit of research and found these answers
Showing that a level set is not a submanifold
How to show a level set isn't a regular submanifold
However I still don't understand, how do I exactly prove that a neighboorhood of the origin is not homeomorphic to any open set in $\mathbb{R}^3$ ? I didn't get the 2 vs 4 components thing
Can you see that the set you are describing is a double cone? Indeed, for any fixed $z$ the equation describes a circle, and as $z$ varies the radii vary. Thus any neighbourhood $U$ of the origin with the origin removed will be not connected. However if $U$ is homeomorphic to some open $V\subset \mathbb R^2$, then $U-\{0\}$ will be isomorphic to $V-\{v\}$ for some point $v$. However the latter is always connected.