Function on $\mathbb{R}^{2}-\{0\}$. Does there exist any compactly supported function $f= (f_1,f_2): \mathbb R^2-\{0\}\to \mathbb R^2$ such that $$\frac{\partial}{\partial x_2}f_1=\frac{\partial}{\partial x_1}f_2.$$
Also there does not exists any function $F: U(\subset \mathbb R^2-\{0\})\to \mathbb R$ such that 
$$\frac{\partial}{\partial x_1}F= f_1\text{ and } \frac{\partial}{\partial x_2}F= f_2.$$
 A: The answer is false, every compact supported $f$ that satisfies $\nabla \wedge f = 0$ is a gradient.
Identify $\mathbb{R}^2$ as $S^2−\{p\}$, the sphere with a point $p$ removed. Let $q$ be the point on $S^2$ that corresponds to $0$ in $\mathbb{R}^2$. Any compact supported vector field $f$ on $\mathbb{R}^2 - \{0\} \approx S^2−\{p,q\}$ vanishes identically on some neighborhood of $p$. As a result, one can extend its definition over the whole of $S^2−\{q\}$.
If $f$ is irrotational, i.e., $\nabla \wedge f = 0$, so is this extension to $S^2−\{q\}$. Since $S^2−\{q\}$ is simply connected, we can find an $F$ on $S^2−\{q\}$ such that $f = \nabla F$. Pull this back to $\mathbb{R}^2 - \{0\}$, and we see any compact supported irrotational vector field on the punctured plane is also a gradient.
Let's look at the problem from another angle without leaving $\mathbb{R}^2$.
For any path or closed loop $\gamma$ in $\mathbb{R}^2 - \{0\}$, 
let $I_{\gamma}(f)$ be the line integral:
$$
I_{\gamma}(f) = \int_{\gamma} ( f_1 dx_1 + f_2 dx_2 )
$$
When $\nabla \wedge f = 0$ and $\gamma$ is a closed loop, Stokes' theorem tells us $I_{\gamma}(f)$ is invariant under continuous deformation of $\gamma$. 
For $f$ with compact support, we can evaluate $I_{\gamma}(f)$ by deforming
$\gamma$ to another loop $\gamma_{\infty}$ outside of $f$'s support. We get:
$$I_{\gamma}(f) = I_{\gamma_{\infty}}(f) = I_{\gamma_{\infty}}(0) = 0$$
Pick an arbitrary point $c \in \mathbb{R}^2-\{0\}$.
For any $x \in \mathbb{R}^2-\{0\}$, let 
$\gamma_i : [0,1] \to \mathbb{R}^2-\{0\}$, $i = 1, 2$
be any two paths that satisfy: $\gamma_i(0) = c, \gamma_i(1) = x$.
Join $\gamma_1$ and $\gamma_2$ to form the closed loop 
$\gamma_{12} : [0,1] \to \mathbb{R}^2-\{0\}$:
$$\begin{align}
\gamma_{12}(t) &= \gamma_1(2t)   & \text{ for } & t \in [0,1/2]\\
          &= \gamma_2(1-2t) &              & t \in [1/2,1]
\end{align}
$$
We have:
$$I_{\gamma_1}(f) - I_{\gamma_2}(f) = I_{\gamma_{12}}(f) = 0$$ 
This means that $I_{\gamma_i}(f)$ is independent of the choice
of $\gamma_i$ and can be viewed as a function $F(c,x)$ of $c$ and $x$ only.
Finally, $F$ satisfies: $\nabla_{x} F(c,x) = f(x)$ and $\nabla_{c} F(c,x) = -f(c)$.
A: Your first equation reduces to $\nabla \wedge f = 0$.  There are plenty of curlless vector fields, so there can exist such a vector field.  This condition says that $f$ is closed.
Your second equations reduce to
$$e_1 \cdot \nabla F = f_1, \quad e_2 \cdot \nabla F = f_2$$
$F$ must be a scalar field for this to be the case, so what you're really saying is that you want,
$$\nabla F = f$$
This may be the case, and by construction it would give $\nabla \wedge f = 0$.  This condition says that $f$ is exact.
Without more information (in particular, the divergence), you can't reconstruct the original vector field $f$.  Since its domain is the punctured plane, it's possible $f$ is closed but not exact--an easy example is $f = e_\theta$.  Of course, the reason this is so is because if you could take the curl at the origin, you would get a nonzero result, even though $f$ would be well-defined (the zero vector) at that point.
Of interest may be Poincare's lemma, which says that closed forms are also exact if, for any two points $p, q$ in the domain, the line segment connecting them is also entirely in the domain (this is not true for the punctured plane).
