$\operatorname{Spec} k[w,x,y,z] / (wz-xy)$ is normal
To recap the discussion from the comments and take this off the unanswered list, the point is you can make a slightly different change of variables to get something more amenable to an application of exercise 5.4.H(a).
In particular, if you change variables by setting $w=(w'-z')$ and $z=(w'+z')$ (this is invertible because $2$ is), we get $w'^2-(z'^2+xy)$, and it is clear that $z'^2+xy$ is irreducible. Therefore we can apply the result of 5.4.H(a) to get that $\operatorname{Spec} k[x,y,z,w]/(wz-xy)\cong \operatorname{Spec} k[x,y,z',w']/(w'^2-(z'^2+xy))$ is normal.