Showing that $F=\frac{1}{x^2+y^2}\langle-y,x\rangle$ is not path independent.

Question

I thought that path independence and being conservative were tied together so using $\displaystyle F=\left\langle{\frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}}\right\rangle$ I found $\dfrac{\partial Q}{\partial x} = \dfrac{y^2-x^2}{(x^2-y^2)^2}$ and then $\dfrac{\partial P}{\partial y} = \dfrac{y^2-x^2}{(x^2-y^2)^2}$, also. But they're the same so that means its is conservative, which doesn't show that it's not path independent. Put them into symbolab to be sure I wasn't making a mistake doing the partial derivatives and got the same.

Answer 1

Consider the line integral around the unit circle. Observe \begin{align} \oint_C F\cdot ds = \int^{2\pi}_0 (-\sin t, \cos t)\cdot (-\sin t, \cos t)\ dt= \int^{2\pi}_0 \ dt = 2\pi \end{align} which means $F$ is not a conservative vector field, i.e. not path independent.

Note: I used the parametrization $c(t) = (\cos t, \sin t)$.