What exactly is a critical point, and what is the relevance of the rank of the Jacobian? Definition
Taken from Wikipedia, the definition of critical points I am used to:
Given a differentiable map $f$ from $\mathbb{R}^m$ into $\mathbb{R}^n$, the critical points of $f$ are the points of $\mathbb{R}^m$, where the rank of the Jacobian matrix of $f$ is not maximal.
I understand this to mean that at a critical point $c \in \mathbb{R}^m$, the derivative $df_c$ is not surjective.

Current understanding
I am used to critical point meaning points in the domain of a single-variable function $f(x)$ where the derivative is zero, and I am struggling to understand the generalisation to multiple variables.
I am particularly struggling with the surjectivity condition. I have seen that surjectivity allows us to "trim" columns of the jacobian and obtain an invertible matrix, which along with the inverse function theorem gives us the implicit function theorem:
given a continuously differentiable function $f:\mathbb{R}^{n+m}\rightarrow\mathbb{R}^{m}$ and a non-critical point $p\in\mathbb{R}^{n+m}$, we can recover the level set $f^{-1}(0)$ in a neighbourhood of $p$ as ($\mathbf{x}, \mathbf{h}(\mathbf{x})$), where $\mathbf{x}$ is $n$ suitable coordinates of $P$.
My question
What exactly breaks down when $df_p$ is not surjective ? Do we lack the information for a suitable linear approximation at p ? Why does the lack of surjectivity tell us something "funny" is happening at $p$ ?

(singular point at (0, 0) of the algebraic curve $y^2-x^3-x^2=0$)
In the above graph, the problem to me seems to be one of injectivity: there are several directions in which $y^2-x^3-x^2$ remains zero.
 A: The function $f$ in your question is $f(x,y)=y^2-x^3-x^2$ from $\mathbb R^2$ to $\mathbb R$ and its Jacobian is the (transpose of the) gradient
$$
(\partial_x f, \partial_y f)=\big(-3x^2-2x,\,2y\,\big)\,.
$$
As a linear map from $\mathbb R^2$ to $\mathbb R$ this is not surjective if and only if it is $(0,0)$, in other words, in $(x,y)=(0,0)$ where the curve defined by
$$
f(x,y)=0
$$
intersects with itself.

*

*Intuitively this happens because a traveller following the curve stays on the level set all the time, and at the critical point has the choice to switch to one of the two branches of the self-intersecting curve.


*At any other point there is a unique direction the traveller has to follow if he or she wants to stay on the level set (curve) because the directional derivative is zero only in that unique direction.


*Work out the relationship between gradient and directional derivative to dive into this. Also (hint): how many orthogonal vectors to the gradient exist at a non-critical resp. at the critical point ?
