Let $v$ be a vector field on $\mathbb{R}^n$. Show that $v$ can be written as a sum $v=f_1\dfrac{\partial}{\partial x_1}+w$ where $w$ is a divergence-free vector field.
Suppose $v=v_1\dfrac{\partial}{\partial x_1}+v_2\dfrac{\partial}{\partial x_2}+\ldots+v_n\dfrac{\partial}{\partial x_n}$, where $v_i:\mathbb{R}^n\rightarrow\mathbb{R}$.
Then we want to choose $f_1$ so that $$v-f_1\dfrac{\partial}{\partial x_1}=(v_1-f_1)\dfrac{\partial}{\partial x_1}+v_2\dfrac{\partial}{\partial x_2}+\ldots+v_n\dfrac{\partial}{\partial x_n}$$ is divergence-free, i.e. this quantity is zero when we actually take these partial derivatives (rather than using them as just symbols).
So, take any point $p\in\mathbb{R}^n$. Then we can evaluate the real number $$q(p)=\dfrac{\partial v_1(p)}{\partial x_1}+\dfrac{\partial v_2(p)}{\partial x_2}+\ldots+\dfrac{\partial v_n(p)}{\partial x_n}$$
If I set $f_1(p)=q(p)x_1$, this yields $v-f_1\dfrac{\partial}{\partial x_1}=0$ at point $p$. Since this holds for any point $p$, we have $v-f_1\dfrac{\partial}{\partial x_1}=0$.
EDIT: This solution is currently wrong, because $q(p)$ is not a constant, so I have to differentiate using the product rule. How can I fix it?
You want $$ \frac{\partial f_1}{\partial x_1}(p) = q(p) .$$ Let $$ f_1(x_1,x_2,\dots,x_n) = \int_0^{x_1} q(\xi,x_2,\dots,x_n) \, d\xi .$$ Here $p = (x_1,x_2,\dots,x_n)$. All is perfectly legitimate because you are working on $\mathbb R^n$. Probably on some other manifolds you could not make this work.