Locally exact form $P\;dx+Q\;dy$ , and the property $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ This is a very known result, but I don't have some proof. Someone known or has some proof of it?

Let be $\omega  = P\;dx + Q\;dy$ be a $C^1$ differential form on a domain $D$. If $$\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} ,$$ then $\omega$ is locally exact. 


If you have a curl-free field $W = (W_1, W_2, W_3)$ in a neighborhood of the origin, it is the gradient of a function $f$ given by
$$ f(x,y,z) =   \int_0^1 \; \left( \;  x W_1(tx, ty,tz) +  y W_2(tx, ty,tz) + z W_3(tx, ty,tz)  \; \right)  dt.$$ 
In your case, take $W_3 = 0$ and drop the dependence on $z$ from $f, \; W_1$ and $W_2.$ Note how this is set up so that $f=0$ at the origin.
There is more information at Anti-curl operator 
