As I know if $$\partial f/\partial y=\partial f/\partial x$$ then f(x,y) is conservative, but there are two counterexamples $$\vec F=\frac{-y\hat\imath+x\hat\jmath}{x^2+y^2}$$ and $$\vec F=\hat\jmath$$
I know there is an similiar question easy question about conservative vector fields
But I still don't understand these two counterexamples contradict with $\partial f/\partial y=\partial f/\partial x$
Can anyone show me how these two counterexamples are non conversative?
thanks
A vector field $\vec F$ is conservative if there exist a function (called potential) $g$ such that $\vec F = \nabla g$. Now it comes out that the integral of $\nabla g$ on any curve $\gamma$ is easily computed as the difference of $g$ on the endpoints of $\gamma$ (this is the fundamental theorem of calculus). So the integral of a conservative vector field along a closed curve is always zero. The converse can be proven to be true, also.
So to prove that a vectorfield $\vec F$ is not conservative it is enough to find a closed curve $\gamma$ on which the integral of $\vec F$ is not zero. In the case of your first example such a curve is any circle centered in the origin.
Your second example $\vec F = j$ is instead an example of a conservative field, since $\vec F = \nabla g$ with $g(x,y,z) = y$.