Consider the vector field $\vec{u}=(xy^2,x^2y,xyz^2)$
The curl of the vector field is $$\nabla \times\vec{u}=(xz^2,-yz^2,0)$$ Consider the line integral of $\vec{u}$ around the ellipse $C$ $x^2+4y^2=1, z=-1$.
With $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a=1, b=\frac{1}{2}$, the parameterisation gives $$\vec{r}=(x,y,z)=(cos\theta,\frac{1}{2}sin\theta,-1)$$ $$d\vec{r}=(-sin\theta,\frac{1}{2}cos\theta,0))$$
$$\vec{u}=(\frac{1}{4}cos\theta sin^2\theta,\frac{1}{2}cos^2\theta sin\theta,\frac{1}{2}sin\theta cos\theta) $$ $$\oint_{C} \vec{u} \cdot d\vec{r}=\frac{1}{4}\int^{2\pi}_0sin\theta cos\theta(cos^2\theta -sin^2\theta)d\theta=0$$ (I got the same result by working in cartesian cooridnates without parameterizing the curve)
But this does not make sense because $$\nabla\times \vec{u}=\lim_{\delta S \to 0}\frac{1}{\delta S}\oint_{\delta C}\vec{u} \cdot d\vec{r}$$ so if a vector field is conservative, its curl should be zero.
Can someone please explain where my conceptual errors lie?
The vector field is not conservative. Being conservative means that FOR EVERY closed curve $C$, we must have $\oint_C \mathbf{u} \cdot \mathbf{dr} = 0$. However, you only checked that for one closed curve (namely that ellipse), the integral vanishes. You didn't check it for every closed curve.
Actually, the very fact that $\nabla \times \mathbf{u} \neq 0$ shows that the vector field is not conservative, because we have the theorem which says \begin{align} \text{ conservative} \implies \text{curl vanishes}, \end{align} so, if we reason contrapositively, we get the equivalent statement \begin{align} \text{curl doesn't vanish $\implies$ not conservative.} \end{align}
Anyway, if you want an explicit counterexample, try to calculate \begin{align} \int_{\gamma}\mathbf{u} \cdot \mathbf{dr} \end{align} where $\gamma:[0,2\pi] \to \Bbb{R}^3$ is the smooth closed curve $\gamma(t) = (\cos t, 1, \sin t)$. You'll find that the integral is strictly positive.