Let $V$ be a vector field on $\mathbb{R}^2$. If $[\frac{\partial}{\partial x},V]=V=[V,\frac{\partial}{\partial y}]$, determine $V$.
The first way to look at this is by observing vector fields primarily as functions from $\mathfrak{F}(\mathbb{R}^2)$ to $\mathfrak{F}(\mathbb{R}^2)$. Obviously, if $D=\frac{\partial}{\partial x}+\frac{\partial}{\partial y}$, then $[D,V]=0$. Writing it out this way: $$\frac{\partial}{\partial x}(V(f))-V(\frac{\partial}{\partial x}f)=V(f)=V(\frac{\partial}{\partial y}f)-\frac{\partial}{\partial y}(V(f))$$, we can also see that $$\frac{\partial}{\partial x}(V(f))=V(f+\frac{\partial}{\partial x}f)$$ $$\frac{\partial}{\partial y}(V(f))=V(\frac{\partial}{\partial y}f-f)$$.
Frankly, I don't know what to do with this. Jacobi identity doesn't give me anything new.
The second way to look at this is to observe these equations locally, at some point $p\in M$. Then we'd be able to write $V$ as $V=\alpha \frac{\partial}{\partial x}+\beta \frac{\partial}{\partial y}$ (or at least I hope we can do this?). However, observed locally, given equations become $0=V=0$, which looks like it's incorrect, because it would mean I have more data than I need.
Other idea was to use the fact that we're working in $\mathbb{R}^2$ and that this is a manifold with one chart basically, but this comes down to the previous idea and the result is the same.
I feel like $[D,V]=0$ is significant here, but I don't know what to do with it. Also, it seems that my attempt at observing things locally is completely flawed. Can we even use the fact that partial derivatives constitute a basis of tangent space the way I've used it?
$\def\del{\partial}$ $\def\ddx{\frac{\del}{\del x}}$ $\def\ddy{\frac{\del}{\del y}}$ Write $V = f\frac{\partial}{\partial x} + g\frac{\partial}{\del y}$. Then for an arbitrary smooth $\phi$ we have \begin{align*} [V, \ddx]\phi &= f\phi_{xx} + g\phi_{xy} - \ddx(f\phi_x + g\phi_y) \\ &= - f_x\phi_x - g_x\phi_y, \end{align*} so $$ [V, \ddx] = -f_x\ddx - g_x\ddy. $$ Similarly, $$ [\ddy, V] = f_y\ddx + g_y\ddy. $$ Both are equal to $V$, so we have $$ -f_x = f_y = f $$ and $$ -g_x = g_y = g. $$ This system of partial differential equations is solved by $f(x, y) = C_1\exp(y-x)$ and $g(x, y) = C_2\exp(y-x)$, so $$ V = C_1\exp(y-x) \ddx + C_2\exp(y-x)\ddy. $$