Let $X$ and $Y$ be a vector field on $M$ and satisfies $[X,Y]=X$.
If $X$ and $Y$ are pointwise linearly independent for some point $p$, then there is a sub manifold $N$ of $M$ such that $T_xN$ is spanned by $X$ and $Y$ by Frobenius theorem.
Are there any counterexample that such $N$ does not exists when $X$ and $Y$ are linearly dependent at point $p$?
OK, here's an example. Let $M=\mathbb R^3$, and define \begin{align*} X = \frac{\partial}{\partial x}, \qquad Y = (x+1) \frac{\partial}{\partial x}+ 3 z^2 \frac{\partial}{\partial y}+2y\frac{\partial}{\partial z} . \end{align*} Then $[X,Y]=X$, and $X$ and $Y$ are linearly independent everywhere except along the $x$-axis. Away from the $x$-axis, the integral manifolds are level sets of $f(x,y,z) = z^3 - y^2$. If there were an integral manifold $N$ containing the origin, by continuity it would have to be contained in the zero-set of $f$; but the zero set of $f$ is not a smooth manifold at the origin because it has a cusp-like fold along the $x$-axis.