The only point of $\operatorname{Spec} (k[x,y] /(y^2,xy))$ with nonreduced stalk is the origin This is exercise 5.2.C in Vakil's lecture notes. I was able to show that $(k[x,y]/(y^2,xy))_x \simeq k[x]_x \simeq k(x)$ is reduced.
Now how does it follow that the only point with nonreduced stalk is the origin ?
If I localize at the origin, I get $(k[x,y]/(y^2,xy))_{(x,y)}$. But I don't see why this is nonreduced.
Progress:
I am still stuck with this exercise.
I cannot take $y \in (k[x,y]/(y^2,xy))_{(x,y)}$, as then $y = 0$ since $xy = 0$. (in the localization). I get the same problem for $x$. There seem to be only constants left ?
 A: As previously noted in the comments, $(k[x,y]/(xy,y^2))_x\cong k[x]_x$ is reduced.
On the other hand, $k[x,y]_{(x,y)}/(xy,y^2)$ is not reduced - $y$ is a nonzero element that squares to zero.
To show that $y\neq 0$ in $k[x,y]_{(x,y)}/(xy,y^2)$, we use the definition of localization: in a ring $R$ with multiplicatively closed subset $S$, the equality $\frac{r}{s}=\frac{r'}{s'}$ holds in $S^{-1}R$ iff there exists a $u\in S$ so that $u(rs'-r's)=0$ in $R$. In our case, we're looking to show that $\frac{y}{1}\neq \frac{0}{1}$.
To make the calculation slightly more understandable, we'll lift to $k[x,y]_{(x,y)}$ - this means we're now on the hook for showing that there's no $u\in k[x,y]\setminus (x,y)$ such that $uy\in (xy,y^2)$. But this is straightforward: an element $u\in k[x,y]$ is not in $(x,y)$ iff it has a nonzero constant term, so we can write $u=c+p$ for $p\in(x,y)$ and $c\in k\setminus 0$. So $uy=cy+yp$ is not in $(xy,y^2)$ because $cy$ is not in $(xy,y^2)$, and so $y\neq 0$ in $k[x,y]_{(x,y)}/(xy,y^2)$.
