Let $X:\mathbb{R}^3 \to \mathbb{R}^3$ be a smooth vector field and $0 \in \mathbb{R}^3$ a regular point (i.e. $X(0)\neq 0$), it is well known that there exists a diffeomorfism $\varphi:V_0 \subset \mathbb{R}^3 \to U_0 \subset \mathbb{R}^3$ ($\varphi(0)=0$), where $U_0, V_0$ are open neighborhood of the origin, such that $$Y(q) = \text{d}\varphi (\varphi^{-1}(q)) \cdot X(\varphi^{-1}(q)) = (1,0,0).$$
The demonstration follows defining $\Phi(x,y,z) = \phi(x,(0,y,z))$ (assunming without loss of generality that $\{X(0),(0,1,0),(0,0,1)\}$ is a basis of $\mathbb{R}^3$), where $\phi(t,z)$ is the solution of the diferencial equation \begin{eqnarray} \left\{\begin{array}{l} \dot{x} = X(x)\\ x(0)=z.\\ \end{array}\right. \end{eqnarray} Defining $\varphi(x,y,z)=\Phi ^{-1} (x,y,z),$ we get that (after some calculations)$$Y(q) = \text{d}\varphi (\varphi^{-1}(q)) \cdot X(\varphi^{-1}(q)) = (1,0,0).$$
A disadvantage of such coordinates is that it is difficult to determine what happens to certain spaces after changing coordinates, for example, the plane $\{0\} \times \mathbb{R^2}$.
Keeping in mind this technical issue I would like to know if this following statement is true and how to prove it
Question: Let $X:\mathbb{R}^3 \to \mathbb{R}^3$ be a smooth vector field, $X = (X_1,X_2,X_3)$ and $0 \in \mathbb{R}^3$ be a point such that $(X_2(0),X_3(0)) \neq (0,0)$.
Is there a diffeomorphism $\varphi: V_0\subset \mathbb{R}^3\to U_0 \subset \mathbb{R}^3$ ($\varphi(0) =0$), such that $\varphi(x,y,z) = (\varepsilon x,f(x,y,z))$ (for some function $f$) and $$Y(q) = \text{d}\varphi (\varphi^{-1} (q)) \cdot X (\varphi^{-1}(q)) = (\varepsilon X_1(\varphi^{-1} (q)),1,0)\ \ \ ?$$ Where $ \varepsilon = \pm 1$.
Can anyone help me?
That seems tricky, but is quite straightforward. Note that the map $\varphi$ you wrote does gives the $Y$ you desire:
Since $X_2(0)\neq 0$, the map $\Phi(x,y,z)=\varphi(y,(x,0,z))$ is invertible at $0$. Note that $d\Phi_q(X(q))=(0,1,0)$. Take $f(x,y,z)=pr\circ \Phi(x,y,z)$, where $pr$ is the projection onto the last two coordinates. You get that $\varphi(x,y,z)=(x,f(x,y,z))$ is invertible at 0: \begin{align} d\varphi_0(1,0,0)&=(1,0,0),\\ d\varphi_0(X(0))&=(X_1(0),1,0),\\ d\varphi_0(0,0,1)&=(0,0,1), \end{align} and
$$d\varphi_q(X(q))=(X_1(q),pr(d\Phi_q(X(q)))=(X_1(q),pr(0,1,0))=(X_1(q),1,0).$$
Summarizing, you just need $X_2(0)\neq 0$ (so that $\{(1,0,0),X(0),(0,0,1)\}$ is a basis).
This right, isn't it? Am I missing something?