How to check there exists a function on given domain whose derivative is given function? I am going through below multiple choice Question, 

Question: let $F_1, F_2: \mathbb{R^2} →\mathbb{R}$ such that, 
  $$F_1(x_1, x_2)=\frac{-x_2}{x_1^2+x_2^2} \text{ and } F_2(x_1, x_2)=\frac{x_1}{x_1^2+x_2^2}.$$ Then which of the following is/are correct?
  
  
*
  
*$\frac{∂F_1}{∂x_2} = \frac{∂F_2}{∂x_1}$ 
  
*there exists function $f: \mathbb{R^2}-\{(0,0)\} →\mathbb{R}$ such that, $\frac{∂f}{∂x_1} =F_1$ and  $\frac{∂f}{∂x_2} =F_2$
  
*there exists No function $f: \mathbb{R^2}-\{(0,0)\} →\mathbb{R}$ such that, $\frac{∂f}{∂x_1} =F_1$ and  $\frac{∂f}{∂x_2} =F_2$
  
*there exists a function $f:D→\mathbb{R}$ where $D$ is open disc of radius 1 centered at $(2,0)$, which satisfies, $\frac{∂f}{∂x_1} =F_1$ and  $\frac{∂f}{∂x_2} =F_2$ on $D$.

My attempt: clearly 
$$\frac{∂F_1}{∂x_2} =\frac{x_2^2 -x_1^2}{\left(x_1^2+x_2^2\right)^2} =\frac{∂F_2}{∂x_1}$$ 
So that (a) is ✓ (correct). But I have no idea about other options. Kindly please help me, facing trouble from hours!! :-(
 A: Let's assume such function $f$ exists, so $\frac{\partial f}{\partial x_1} = F_1$. Then,
$$
f(x_1, x_2)
 = \int \frac{-x_2 dx_1}{x_1^2 + x_2^2}
 = -\arctan(x_1/x_2) + C(x_2)
$$
Thus,
$$
\frac{\partial f}{\partial x_2} = \ldots
$$
Can you compute the partial derivative, set it equal to $F_2$ and see if it matches, and then finish the problem?
UPDATE
Differentiating, we get
$$
\frac{x_1}{x_1^2+x_2^2}
    = F_2
    = \frac{\partial f}{\partial x_2}
    = \frac{x_1}{x_1^2+x_2^2} + C'(x_2)
$$
which implies $C'(x_2) = 0$, so $C$ is a constant, not depending on $x_2$. Thus,
$$
f(x_1, x_2)
 = -\arctan(x_1/x_2) + C
$$
seems to satisfy the algebraic conditions of the problem. But notice this function is not defined anywhere at $x_2=0$. Can you finish the problem now?
A: The vector field ${\bf F}=(F_1,F_2)$ is nothing else but $\nabla{\arg}$, where ${\rm arg}(x,y)$ denotes the polar angle of $(x,y)$, modulo $2\pi$. With this in mind it becomes clear that only 2) is false. But we need a proof of this not making use of the "background information" given here. 
In order to prove that the field ${\bf F}$ has no potential $f$, even though its curl is $\equiv0$, we need a stronger tool: Integrate ${\bf F}$ along the unit circle
$$\gamma:\quad t\mapsto (\cos t,\sin t)\qquad(0\leq t\leq2\pi)\ .$$
You obtain
$$\int_\gamma {\bf F}\cdot d{\bf z}=\int_0^{2\pi}\bigl((-\sin t)(-\cos t)+\cos t\cos t\bigr)\>dt=2\pi\ .$$
Since this integral is $\ne0$ the field ${\bf F}$ cannot have a potential.
