I know that for a continuously differentiable vector field $F: D \subset R^3 \to R $ where $D$ is an simply connected subset of $R^3$ to be conservative, it must satisy $curl(F)(P) = 0$ at every $P \ \epsilon \ D$.
But $curl(F)(P) = 0$ for every $P \ \epsilon \ D$ is a necessary condition for the vector field to be consevative only when the vector field is continuously differentiable. So what I want to ask is the follwing:
Does there exist a conservative vector field which is differentiable but not continuously differentiable and whose $curl \ne 0$ at every point in its domain? If not can we show that a differentiable vector field must always have a vanishing curl?
I think you are referring to counterexamples in the context of Clairaut's theorem, i.e., examples of functions where the order of partial derivatives may not be interchanged. Wikipedia has an example in two variables with a nice picture.
One could adapt such an example to establish nonvanishing curl at a single point: add a dummy third variable $z$ and consider the gradient of that function. A gradient is always conservative, but the $z$-component of the curl of that gradient is nonzero at the origin.
Such examples do not work for constructing a nonvanishing curl in the entire domain, however; distribution theory shows that partial derivatives can be freely interchanged up to a set of volume 0.