Stewart in "Multivariate Calculus" defines a conservative vector field as a gradient of some scalar function. He goes on to say that a conservative vector field has zero curl.
How about $$f(x, y, z) = \begin{cases} \frac{xy(x^2-y^2)}{x^2+y^2} &\mbox{if } (x,y,z) \neq 0 \\ 0 & \mbox{if } (x,y,z)=0, \end{cases}$$ which according to https://calculus.subwiki.org/wiki/Failure_of_Clairaut's_theorem_where_both_mixed_partials_are_defined_but_not_equal has second partials on all of $\mathbb{R}^3$ and satisfies $f_{xy} (0,0,0) \neq f_{yx} (0,0,0)$? If I'm not mistaken, won't the gradient of this function have a non-zero curl at $(0,0,0)$?