From Intro to Smooth Manifolds by Lee:
The map $\gamma: \mathbb R \to \mathbb R^2$ given by $\gamma(t)=(t^3,0)$ is smooth and a topological embedding, but it is not a smooth embedding because $\gamma'(0)=0$.
I know this map is not a smooth embedding because it is a not an immersion since $\gamma'(t)=(3t^2,0)$ and thus $\gamma'(1)=\gamma'(-1)$ and hence the derivative is not injective.
How does $\gamma'(0)=0$ imply that the map is not a smooth immersion?
Expanding Travis' comment, at any point $t$, the differential of $\gamma$ is $d\gamma_t: T_t \mathbb R \to T_{\gamma(t)} \mathbb R^2$ defined as $$\begin{align*} \gamma'(t)&=d\gamma_t\left(\frac{d}{dt}\bigg|_t\right)\\ &=\frac{d\gamma^1}{dt}(t) \frac{\partial}{\partial x^1}\bigg|_{\gamma(t)}+\frac{d\gamma^2}{dt}(t) \frac{\partial}{\partial x^2}\bigg|_{\gamma(t)}\\ &=3t^2 \frac{\partial}{\partial x^1}\bigg|_{\gamma(t)}+ 0 \frac{\partial}{\partial x^2}\bigg|_{\gamma(t)} \end{align*}.$$
So, the matrix of $d\gamma_t$ is $$\begin{pmatrix} 3t^2 \\ 0 \end{pmatrix}.$$
When $t=0$, the differential $d\gamma_0$ is the zero transformation which is not injective. So, $\gamma$ is not a smooth immersion.