Smooth manifold and singular point I came across a statement that looks like this: 
Let $f(x): x\in U\subseteq\mathbb{R}^d\rightarrow\mathbb{R}$ be an analytic function. Then, $\left\{x\in U:f(x)=y\right\}$ is a smooth manifold if $\nabla f(x)\neq0$ for all $x$ such that $f(x)=y$.
Why is this the case? Why can't smooth manifolds include points that is $\nabla f(x)=0$?
 A: This an example of the Implicit Function Theorem.
Let $M = \left\{x\in U \mid f(x) = y\right\}$, and let $a \in U$.  Since $\nabla f(a) \neq 0$, one of the partial derivatives $\frac{\partial f}{\partial x_i}(a) \neq 0$.  Let's assume it's $\frac{\partial f}{\partial x_d}(a)$.  The Implicit Function Theorem says that there exists an open set $V$ in $\mathbb{R}^{d-1}$, containing the point $(a_1,\dots,a_{d-1})$, and a function $g \colon V \to \mathbb{R}$, such that $g(a_1,\dots,a_{d-1}) = a_d$, and 
$$
    f(x_1,x_2,\dots,x_{d-1},g(x_1,x_2,\dots,x_{d-1})) = y
$$
That is, every point of $M$ has a neighborhood where one of the coordinates can be written as a function of the others.  If $f$ is $C^k$, then $g$ is $C^k$ too.  The derivatives of $g$ can be computed in terms of the derivatives of $f$.
As an example, let $f(x_1,x_2) = x_1^2 + x_2^2$, $y=1$, and $a = (0,1)$.  Near $a$, we can write $x_2 = \sqrt{1-x_1^2}$, and the function $g(x) = \sqrt{1-x^2}$ is smooth in a neighborhood of $0$.  This tells us that the single coordinate $x_1$ can describe a point on the circle in a neighborhood of $(1,0)$.  This won't work near $(1,0)$, though, and notice $\frac{\partial f}{\partial x_2}(1,0) = 0$.  So we use the other coordinate $x_1$, and write it in terms of $x_2$.
As a nonexample, consider $f(x_1,x_2) = (x_1^2 + x_2^2)^2 - (x_1^2-x_2^2)$.  The level set $f^{-1}(0)$ is called a lemniscate, and looks like this:

You can see that except for the origin, the curve is nice and smooth.  But at the origin, the curve intersects itself transversely.  In terms of $\nabla f$, we have,
\begin{align*}
    \frac{\partial f}{\partial x_1} &= 2(x_1^2 + x_2^2)(2x_1) - 2x_1 
        = 2x_1(2(x_1^2 + x_2^2)-1) 
\\
    \frac{\partial f}{\partial x_2} &= 2(x_1^2 + x_2^2)(2x_2) + 2x_2 
        = 2x_2(2(x_1^2 + x_2^2)+1) 
\\
\end{align*}
So $\nabla f(x_1,x_2) = 0 \iff (x_1,x_2) = (0,0)$.
The basic idea of a manifold is a topological space where every point has a neighborhood that “looks like” a subset of $\mathbb{R}^n$.  In this setting, a given a point $a \in M$, there is a subset $V$ of $\mathbb{R}^{d-1}$ and a map $\phi \colon V \to M$, given by
$$
    \phi(x_1,\dots,x_{d-1}) = (x_1,\dots,x_{d-1},g(x_1,\dots,x_{d-1})
$$
Let $V' = \phi(V)$; then $a \in V' \subset M$.  Then $V'$ looks like $V$, in the sense that we can use coordinates on $V$ to describe points of $M$.  That is why $M$ is a smooth manifold.  
To learn more about manifolds and how they are defined and analyzed, look for books on “Differential Geometry” or “Differential Topology”.  Boothby is the one that I learned from, but volumes by do Carmo or Guillemin and Pollack are also good.
