Integral depending on a path? I need to check whether differential form $\omega$ has, in the domain $G$, such property that it's integral doesn't depend on path. In my exercise: $\omega = \frac{ydx -xdy}{x^2+xy+y^2}$ and $G= R^2 \setminus {(0,0)}$.
So I found the antiderivative, it's equal to $\frac {2}{\sqrt{3}}\arctan\left(\frac{2x+y}{\sqrt{3}y}\right)$. And my problem is that it isn't defined for $y=0$. So what should I do now to answer my question? Define it for $y=0$? Or maybe it will turn out that this integral depends on a path?
It is true that your differential
$$\omega=Pdx+Qdy:={y\over x^2+xy+y^2}\ dx-{x\over x^2+xy+y^2}\ dy$$
satisfies the integrability condition $Q_x-P_y\equiv0$. Therefore in simply connected subsets $\Omega\subset G$ there exist potential functions $f:\ \Omega\to{\mathbb R}$ with $df =\omega$, or $\nabla f=(P,Q)$.

But there is no global $f:\ G\to{\mathbb R}$ of this sort. For a proof consider the elliptic curve
$$\gamma:\quad t\mapsto\bigl(x(t),y(t)\bigr):=\left(\cos t-{1\over\sqrt{3}}\sin t,\ {2\over\sqrt{3}}\sin t\right)\qquad(0\leq t\leq2\pi)\ .$$
One computes
$$x^2(t)+x(t)y(t)+y^2(t)\equiv1,\quad \dot x(t)=-\sin t-{1\over\sqrt{3}}\cos t,\quad \dot y(t)={2\over\sqrt{3}}\cos t\ .$$
It follows that
$$\int_\gamma \omega=\int_0^{2\pi}\bigl(y(t)\dot x(t)-x(t)\dot y(t)\bigr)\ dt=-{4\pi\over\sqrt{3}}\ne0\ .$$
You can write your $\omega$
$$\omega=\frac{y}{x^2+xy+y^2}dx-\frac{x}{x^2+xy+y^2}dy$$
For integral it has the form
$$\int\bigg(Pdx+Qdy\bigg)$$
which you can test for beeing conservative such as
$$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$$
where
$$\frac{\partial P}{\partial y}=\frac{(x-y)(x+y)}{\big(x^2+xy+y^2\big)^2}$$
$$\frac{\partial Q}{\partial x}=\frac{(x-y)(x+y)}{\big(x^2+xy+y^2\big)^2}$$
Which means that the field is conservative and the line integral is path-independent.

EDIT

I must have said in the beginning "Assuming that the path is closed..."

and at the end "This is a necessary requirement but not sufficient"