The set $N = \left\lbrace \left(x,y\right)\in\mathbb R^2\mid x^2-y^2-x^4=0\right\rbrace$ is given and I want to prove that $N$ is not a submanifold of $\mathbb R^2$. Our definition of a subset $N\subseteq \mathbb R^n$ being a $k$-dimensional submanifold is that for every $x\in N$ there exists an open neighborhood $U_x\subseteq\mathbb R^n$ and a diffeomorphism $\phi:U_x\to \phi(U_x)=:V_x$ with $V_x\subseteq \mathbb R^n$ also being open so that $$\phi(N\cap U_x)=V_x\cap\left(\mathbb R^k\times\{0\}^{n-k}\right)=\left\lbrace y\in V_x \mid \forall i>k:y_i=0\right\rbrace.$$ From this definition it follows readily that an $n$-dimensional submanifold of $\mathbb R^n$ is open and a $0$-dimensional one is a discrete set. Both of these are not true for my set $N$, so the only remaining option would be a one-dimensional submanifold.
Looking at the plot of this curve on WolframAlpha I think the "problematic" point is $(0,0)$, but I don't know how to prove that there exists no such diffeomorphism at this point. How should I do that?
Bonus question: How can I find these problematic points and get an idea of what the curve looks like just from the equation given and without WolframAlpha (i. e. in an exam)?
Yes, if it were going to be a submanifold, it would have to be $1$-dimensional. To see locally what it looks like (without Wolfram), notice that near the origin it looks like $x^2-y^2=0$, i.e., an X. If it fit your definition, after some suitable change of coordinates, a neighborhood of the origin would look like $y=0$ sitting inside a neighborhood of the origin in $\Bbb R^2$. Can you see what that's not possible?
The Implicit Function Theorem guarantees you that you will have a submanifold locally around any point where the derivative of your defining function is nonzero. So the problem points come from critical points of that defining function.
For future reference, you might want to look at the answer to this question.