Prove $\lim_{z \to 0} \frac{z}{\overline{z}}$ doesn't exist using $\varepsilon-\delta$.

The epsilon-delta argument can be made very simply, once you know that the limiting value is path-dependent. Let $$f(z) = z/\bar z = e^{2i\arg(z)}.$$ Then suppose there exists an $L \in \mathbb C$ satisfying the definition; then the claim the limit exists is equivalent to stipulating that $|e^{2i \arg(z)} - L|$ can be made arbitrarily small when $z$ is in a neighborhood of $0$. But you can see right away where this will not work: the magnitude of $e^{2i \arg (z)}$ is always unity irrespective of the size of the neighborhood, but the argument is $2 \arg (z)$; thus if you choose any fixed $L$, the supremum of the modulus of the difference is never less than unity. Geometrically, this is equivalent to saying that for any choice of a point in the plane, the maximum distance of that point to any point on a unit circle is never less than $1$. This furnishes the intuition for proceeding with a more formal argument, the outline of which is as follows:

We may assume without loss of generality that $\Im(L) = 0$ and $\Re(L) \ge 0$. Then we compute for such an $L$ the maximum value of $|f(z) - L|$, which occurs for $\arg(z) = \pm \pi/2$; hence $|f(z) - L| = L+1$, and it follows that for any choice of $\epsilon < 1$, it is impossible to choose $\delta > 0$ such that whenever $|x| < \delta$, $|f(z) - L| < \epsilon$.

With $z=e^{i\theta}$ we have

$$\frac z{\bar z}=e^{2i\theta}=\cos2\theta+i\sin2\theta,$$ independently of $r$.

Then as

$$\left|\cos2\cdot0-\cos2\frac\pi2\right|=2,$$ for $\epsilon<1$, no $\delta$ can satisfy the condition

$$|f(z)-L|<\epsilon.$$