= Solution
For a <ring homomorphism> $A\to B$, the <Module of Kähler differentials> $\Omega_{B/A}$ is the $B$-module generated by symbols $db$, subject to
$$
d(b+b')=db+db',\qquad d(bb')=b\,db'+b'\,db,\qquad da=0\quad(a\in A).
$$
Equivalently, it represents $A$-linear <derivation of an algebra>[derivations]:
$$
\operatorname{Hom}_B(\Omega_{B/A},M)\simeq\operatorname{Der}_A(B,M).
$$
For $A\to B\to C$, the <Transitivity exact sequence for Kähler differentials> is
$$
\Omega_{B/A}\otimes_BC\longrightarrow\Omega_{C/A}\longrightarrow\Omega_{C/B}\longrightarrow0.
$$
If $B\to C=B/I$ is surjective, the <Conormal exact sequence for Kähler differentials> is
$$
I/I^2\longrightarrow\Omega_{B/A}\otimes_BC\longrightarrow\Omega_{C/A}\longrightarrow0,
\qquad [f]\longmapsto df\otimes1.
$$
Let $L/K$ be a finite <field extension>. By the <primitive element theorem>, its maximal <separable field extension> $L_s/K$ is simple, and transitivity reduces the calculation to a simple algebraic extension. If $L=K(\alpha)$ with <minimal polynomial> $m_\alpha$, then
$$
\Omega_{L/K}\simeq L\,d\alpha/(m_\alpha'(\alpha)d\alpha).
$$
Thus a separable simple extension has zero differentials. Conversely, if $L/K$ is not separable, the purely inseparable part has a generator whose minimal polynomial has zero <formal derivative in positive characteristic>, producing a nonzero differential. Hence
$$
\boxed{\Omega_{L/K}=0\iff L/K\text{ is separable}.}
$$
Now let $\operatorname{char}K=p>0$. If $L=K(a)$, $a^p\notin K$, and $a^{p^2}\in K$, then the minimal polynomial is $T^{p^2}-a^{p^2}$ and has zero derivative, so
$$
\boxed{\Omega_{L/K}=L\,da.}
$$
If $L=K(a,b)$, where $a\notin K$, $b\notin K(a)$, and $a^p,b^p\in K$, then $[K(a):K]=[L:K(a)]=p$. Both defining equations have zero derivative, and
$$
\boxed{\Omega_{L/K}=L\,da\oplus L\,db.}
$$
For the three morphisms, compute the <Sheaf of relative Kähler differentials> on coordinate rings.
* In (i), $x=y^2$ and $dx=0$ relative to $k[t]$, so
$$
\Omega_{X/Y}\simeq k[y],dy/(2y\,dy).
$$
If $\operatorname{char}k\ne2$, this is $k[y]/(y)\,dy$ and its <support of a module>[support] is the origin $V(y)=V(x,y)$. If $\operatorname{char}k=2$, it is free of rank one and its support is all of $X$.
* In (ii), $d(x^2)=2x\,dx$ and $d(x^3)=3x^2\,dx$ vanish relatively, whence
$$
\Omega_{X/Y}\simeq k[x],dx/(2x,3x^2)dx.
$$
Its support is the origin $V(x)$ in every characteristic; the module is $k[x]/(x)$ unless $\operatorname{char}k=2$, when it is $k[x]/(x^2)$.
* In (iii), $dx=0$ and $d(xy)=y\,dx+x\,dy$ give
$$
\Omega_{X/Y}\simeq k[x,y]/(xy,x)\,dy\simeq k[y],dy.
$$
Its support is the entire component $V(x)\subset X$.
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