= Solution
A <morphism of schemes> $f:X\to Y$ is a <flat morphism> when every local-ring map $\mathcal O_{Y,f(x)}\to\mathcal O_{X,x}$ makes $\mathcal O_{X,x}$ a <flat module>.
In (i), the coordinate map is $k[t]\to k[y]$, $t\mapsto y^2$, and
$$
k[y]=k[t]\oplus yk[t].
$$
It is therefore a <free module> of rank two and the morphism is flat, including in characteristic two.
In (ii), $k[x]$ is finite over the cusp ring $R=k[x^2,x^3]$ and has generic rank one. Were it flat, finite flatness over the local ring at the cusp would make it free of rank one. Its fiber there is instead
$$
k[x]\otimes_R R/(x^2,x^3)\simeq k[x]/(x^2),
$$
which has dimension two, so this morphism is not flat.
In (iii), the base coordinate $t$ acts as $x$, and the nonzero element $y$ satisfies $ty=xy=0$. Thus the coordinate ring has torsion as a $k[t]$-module. Since $k[t]$ is a <principal ideal domain> and a module over it is flat exactly when it is torsion-free, this morphism is not flat. Consequently
$$
\boxed{\text{only (i) is flat}.}
$$
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