Let $f:\mathbb{R}\to\mathbb{R^{2}}$, $f(t)=(t^2, t^3)$, then the image of $f$ is not a submaniflod.
The definition of submanifold I learned is the following (some others call this regular submanifold):
A subset $S\subseteq M$ is called a submanifold of dimension $k\leq m$, if it has the following property: For every $p\in S$ there is a coordinate chart $(U, \varphi )$ around $p$, such that $\varphi (U\cap S)= \varphi (U)\cap\mathbb{R^k}$.
I know that if $F$ is an embedding, then image of F is a regular submanifold, but here f is not an embedding, so can not use this theorem.
And also, there is another answer https://math.stackexchange.com/a/1090680/1141262, but I don't understand how this answer use $D\alpha (0)=0$ to show that no chart $(U, \varphi )$ around $p$ exists, such that $\varphi (U\cap S)= \varphi (U)\cap\mathbb{R^k}$
Edited: I kind of understanding why from the above definition one can show submanifold is also locally a graph.
And I saw an answer, but I don't understand how the answer showed the image is not an submanifold. Showing that $\tau(t) = (t^2, t^3)$ is not a submanifold I think if a set is a graph of a smooth fucntion, then the set is a manifold, but I cannot understand why being a submanifold need locally to be a graph of a function as showed in the answer.