Which level sets of a function are submanifolds I want to determine which of the level sets $f^{-1}(c)$ of $f\colon \mathbb{R}^2\longrightarrow \mathbb{R}, \  (x_1,x_2)\mapsto -x_1^3+x_1^2+x_2^2$ for every $c\in\mathbb{R}$, are submanifolds.
Firstly, I computed $\nabla f(x_1,x_2)=(-3x_1^2+2x_1,2x_2)^t$.
The critical points of $\nabla f$ are $(3/2,0),(0,0)$.
For all other points the nabla of $f$ is not the zero vector. I then determined the images of these points, which are $(-27/8,0)$ and $(0,0)$. Finally, I concluded that $\mathbb{R}^2$ without these images are all submanifolds of dimension 1.
Is this correct?
EDIT: I somehow added another dimension to $f$ before, this is gone now.
 A: There are differential topology theorems that answer this question.
Consider the preimage/submersion/regular value theorem.
Also related to it is a celebrated theorem by Sard.
Sard implies that almost every $q\in \text{Im}(f)$ will be a regular value for $f$, which implies via the Preimage Theorem that $f^{-1}(q)$ is a submanifold of $\text{Dom}(f)$.

For your specific example, let's check which $c\in\mathbb{R}$ are regular values of $f$.
The differential $Df_p:\mathbb{R}^2\longrightarrow \mathbb R$ is the linear map
$$Df_p(v)=\langle \nabla f(p), v\rangle.$$
Hence, in order for $Df_p$ to fail to be surjective it must be that $\nabla f(p)=0$.
We calculate $\nabla f$ at $p=(x,y)$.
We have
\begin{align}
\nabla f(x,y)
&=
\left(\frac{\partial f}{\partial x_1}(x,y), \frac{\partial f}{\partial x_2}(x,y)\right)\\
&=(-3x^2+2x,2y)
\end{align}
and hence $\nabla f(x,y)=0$ if and only if $y=0$ and $-3x^2+2x=0 \iff x = 0$ or $x=2/3$.
In other words, our critical points are $(0,0)$ and $(2/3,0)$.
We have $f(0,0)=0$ and $f(2/3,0)=-8/27+4/9=4/27$, so these are our critical values.
For all $c\in\mathbb{R}\setminus\{0,4/27\}$ the Preimage Theorem then guarantees that $f^{-1}(c)$ will be either empty or a differentiable submanifold of $\mathbb{R}^2$, possibly disconnected, with dimension $1$.
It's easy to see that $f$ is surjective, so $f^{-1}(c)$ is never empty, which concludes the exercise.
