Let $A=k[x,y]/(x^2,xy)$. Consider $Ass(A/0)$ as in the previous proposition. We have one possible primary decomposition of $0$:

Notice that $(\overline{y})$ is not prime here, since ${\overline{x}}^2=0\in (\overline{y})$, but $\overline{x}\notin (\overline{y})$. Therefore $r(\overline{x})=(\overline{x})$. Also notice that $(\overline{y})$ is primary. If a polynomial $fg\in (\overline{y})$, and $f\notin (\overline{y})$, then that means that $f$ has an $\overline{x}$ term or a constant term. If $g$ also had a constant term, then $fg\notin (\overline{y})$. The only possibility is that $g$ has an $x$ term, but then $g^2\in (\overline{y})$, since $x^2=0$ and $xy=0$.

Thus $Ass(A/0)=\{(\overline{x}),(\overline{x},\overline{y})\}$, and we see that the nilradical is $(\overline{x})\bigcap (\overline{x},\overline{y})=(\overline{x})$, and the zero divisors are $(\overline{x})\bigcup (\overline{x},\overline{y})=(\overline{x},\overline{y})$.