Smooth Coordinates for which a Vector Field is a Coordinate Vector Field Given a vector field $W = x\frac{\partial}{\partial x}+2y\frac{\partial}{\partial y}$, I want to find smooth coordinates around $(1, 0)$ such that this vector field is a coordinate vector field. I know that the flow for this field is given by $\theta_t(x,y)=(xe^t,ye^{2t})$.
I think that I need to find a smooth coordinate representation of $W$ such that mapping $(1,0)\mapsto \frac{\partial}{\partial x^i}\big|_{(1,0)}$ gives us our original field, but I have no clue explicitly how to calculate it. I think I need to calculate flows for the representation (probably parameterized by other variables), but I'm not sure. If it helps, this is problem 9-10 from Lee's Intro to Manifolds.
 A: In Lee's book, take a close look at Theorem 9.22, which has the label "Canonical Form Near a Regular Point" (it is on page 220 in the edition I'm using). It is the theorem which states that what you are trying to do is possible (i.e. find coordinates so that the chosen vector field is a coordinate vector field). This is also often called the "straightening theorem" (e.g. this Wikipedia page)
If you read the proof, it tells you how to construct such coordinates.
Start by finding a curve through the point $(1,0)$ which is not tangent to the vector field near the point. For example, we can take the vertical line $x=1$. We can parameterize this line using the $y$-coordinate. For each point on this vertical line, we can apply the flow of $W$ for time $t$ (for $t \in (-\varepsilon, \varepsilon)$), and this will generate a small neighborhood around $(1,0)$.
In other words, our new coordinate system is $(t,v)$, which is the point obtained by starting at $(1,v)$ (i.e. a point on the vertical line, in the original $x,y$ coordinates) and flowing for time $t$, using the $\theta_t$ you mentioned. This gives the coordinate transformation
$$ (x,y) = (e^t, v \, e^{2t}) $$
You can confirm (using the chain rule) that $\frac{\partial}{\partial t}$ is the desired vector field:
$$ \frac{\partial}{\partial t} = \frac{\partial x}{\partial t} \frac{\partial}{\partial x} + \frac{\partial y}{\partial t} \frac{\partial}{\partial y} = e^t \, \frac{\partial}{\partial x} + 2v e^{2t} \, \frac{\partial}{\partial y} = x \, \frac{\partial}{\partial x} + 2y \, \frac{\partial}{\partial y} $$
