Given a vector field $F$ in polar coordinates, for the example the field $$\vec F(r,\theta)= -r \hat r + (r^2\sin \theta ) \hat \theta $$
I am asked to check if the field is conservative. is it right to say that the curl of the field is not $0$, because $\frac d {dr} (r^2 \sin \theta ) - \frac d {d\theta} (-r) = 2r\sin \theta \neq 0$, and therefore the field is not conservative? Or do I have to change first the coordinates to cartesian coordinates and then do the same/other process? ( I had this question in an exam and the lecturer changed to cartesian coordinates before he made this process in his exam solution he published... )
No, your approach is not correct. In polar coordinates the curl of $F$ is (see this page): $$\frac{1}{r}\left(\frac{\partial (r F_{\theta})}{\partial r}-\frac{\partial F_{r}}{\partial \theta}\right)=\frac{1}{r}\left(\frac{\partial (r^3\sin\theta)}{\partial r}-\frac{\partial (-r)}{\partial \theta}\right)=3r\sin\theta.$$ Hence the field is not conservative in any open set of $\mathbb{R}^2$.
The same result can be found using cartesian coordinates: $$F=-r \hat r + (r^2\sin \theta ) \hat \theta=(-x-y^2)\hat i+(-y-xy)\hat j$$ and therefore the curl of $F$ is $$\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}=\frac{\partial (-y+xy)}{\partial x}-\frac{\partial (-x-y^2)}{\partial y}= y+2y=3y$$ which coincides with the previous computation.