Here is exercise 1 from chapter 8.1 of Bosch - Algebraic Geometry and Commutative Algebra:
(Exercise:)
For a field $K$, consider the coordinate ring $A = K[t_1, t_2]/(t_2^2 − t_1^3)$ of Neile’s parabola. Show that the $A$-module of relative differential forms $\Omega_{A/K}^1$ can be generated by two elements, but that it is not free. Conclude once more that the scheme $Spec \, A$ cannot be isomorphic to the affine line $\mathbb{A}_K^1$.
I'm posting my solution below and I will be very grateful if someone can check whether it's correct or not. I think my solution is a bit too long, so it would also be interesting to have other solutions using modules of relative differentials (I know we can prove the same result by showing that $A$ is not integrally closed but the purpose here is to use modules of differentials).
(My solution)
Let $A = K[x,y]/(y^2 - x^3)$. Define $X = Spec \, A$ and denote $O$ the point in $\mathbb{A}_K^2$ related to the maximal ideal $(x,y) \subset K[x,y]$. We can see $X$ as a closed subscheme of $\mathbb{A}_K^2$ containing $O$.
Let $B = K[t]$. Then every morphism $\psi: \mathbb{A}_K^1 \to X$ is categorically equivalent to a morphism $\psi^{\sharp}: A \to B$ of $K$-algebras.
Fix $\overline {K}$ an algebraic closure of $K$. Applying the functor $- \otimes_K \overline{K}$ to the category of $K$-algebras provides a canonical way to obtain a morphism $\psi_{\overline{K}}: \mathbb{A}_{\overline{K}}^1 \to X \times_K \overline{K}$ from a morphism $\psi: \mathbb{A}_K^1 \to X$. Since $\overline{K}$ is a flat $K$-module (every extension of $K$ is), we have that: $\psi$ is an isomorphism $\Rightarrow \psi_{\overline{K}}$ is an isomorphism.
We will show that a morphism $\psi: \mathbb{A}_K^1 \to X$ cannot be an isomorphism by proving that there is no isomorphism $\mathbb{A}_{\overline{K}}^1 \to X \times_K \overline{K}$. Up to replacing $\psi$ by $\psi_{\overline{K}}$ and $\psi^{\sharp}$ by $\psi_{\overline{K}}^{\sharp} :=\psi^{\sharp} \otimes_K Id_{\overline{K}}$ we can assume from now on that $K$ is an algebraically closed field.
Any non-trivial morphism of $K$-algebras $A \to B $ fits in the following general form: $\begin{cases} A \to B \\ \overline{x} \mapsto P(t) \\ \overline{y} \mapsto Q(t) \end{cases}$ where $P,Q$ are non-zero polynomials in $K[t]$ satisfying $Q^2 = P^3$. Note that $K$ algebraically closed $\Rightarrow K$ infinite, hence we can identify polynomials to polynomial functions.
The morphism $\psi^{\sharp}: A \to B$ yields the follwoing canonical exact sequence of $B$-modules:
$$\Omega_{A/K}^1 \otimes_A B \xrightarrow{f_1} \Omega_{B/K}^1 \xrightarrow{f_2} \Omega_{B/A}^1 \to 0$$
Its is clear that $\Omega_{B/K}^1 \cong Bdt$. We can also show that $\Omega_{A/K}^1 \cong (Ad\overline{x} \oplus Ad\overline{y})/(2\overline{y}dy - 3\overline{x}^2dx)$. The morphism $f_1$ is explicitely given by $$ f_1 : \begin{cases}\Omega_{A/K}^1 \otimes_A B \to \Omega_{B/K}^1 \\ d \overline{x} \otimes 1 \mapsto P'(t)dt \\ d \overline{y} \otimes 1 \mapsto Q'(t)dt \end{cases}$$
Suppose that $\psi^{\sharp}: A \to B$ is an isomorphism then we must have $\Omega_{B/A}^1 = 0$ and $f_1$ must be an isomorphism. We will get a contradiction by showing that $f_1$ cannot be an isomorphism.
In general, the Nullstellensatz implies that $z \in \mathbb{A}_K^1$ is a closed point $\iff \kappa(z)$ is a finite extension of $K$. Since $K$ is assumed to algebraically closed: $z \in \mathbb{A}_K^1$ is a closed point $\iff \kappa(z) = K$
Let $z \in \mathbb{A}_K^1$ be any point lying in the fiber $ \psi^{-1}(O)$. Since $O$ is a closed point, so is $z$. Let $t_{z} \in \kappa(z) = K$ be the evaluation of $t \in \Gamma(\mathcal{O}_{ \mathbb{A}_K^1}, \mathbb{A}_K^1)$ at the point $z_O$. Its clear that $P(t_{z} ) = Q(t_{z} ) = 0$. Conversely if $\alpha \in K$ is a common root of $P$ and $Q$, then $\alpha$ corresponds canonically to a point $z \in \mathbb{A}_K^1$ such that $t_{z} = \alpha$. Since $P(t_{z} ) = Q(t_{z} ) = 0$, we obtain that $z \in \psi^{-1}(O)$.
We have just shown that the fiber $ \psi^{-1}(O)$ is in bijection with the common roots of $P$ and $Q$. Since $\psi$ is assumed to be an isomorphism, the fiber $\psi^{-1}(O)$ has only one point. This shows that $P$ and $Q$ have only a single common root in $K$ which we will denote by $\alpha$. Let us use this fact to show that $dt$ does not belong to $Im(f_1)$:
It is clear that every $\omega \in Im(f_1)$ takes the form $\omega = [G(t)P'(t) + H(t)Q'(t)]dt$ with $G,H$ polynomials in $ K[t]$. If we impose $G(t)P'(t) + H(t)Q'(t) = 1$ in order to have $\omega = dt$, we obtain by Bezout's Identity in $K[t]$ that $P'$ and $Q'$ must be coprime.
In $K$, the roots of $P^3$ are exactly the same roots of $P$ with three times the multiplicity. The same argument works for $Q^2$ and $Q$. Since $Q^2 = P^3$, we obtain that $P$ and $Q$ have exactly the same roots in $K$ (without taking multiplicity into account). Under the assumption that $\psi$ is an isomorphism, we have shown that $P$ and $Q$ have only a single common root $\alpha$. Thus $P(t) = (t - \alpha)^n$ and $Q(t) = (t - \alpha)^m$ with $n, m$ integers $\geq 1$.
Since $Q^2 = P^3$, we have that $2 Q(t)Q'(t) = 3 P^2(t) P'(t)$. Multiplying both terms by $P(t)$ and then replacing $P^3(t)$ by $Q^2(t)$ in the RHS. We obtain: $2P(t)Q'(t) = 3 Q(t)P'(t)$.
If $char(K) \neq 2,3$ define: $S_{P'} \subset K$ the set of roots of $P'$ and $S_{Q'}$ the set of roots of $Q'$.Let $ \beta \in S_{P'}$. The equality $2P(t)Q'(t) = 3 Q(t)P'(t)$ implies $P(\beta)Q'(\beta) = 0$. Since $P',Q'$ are coprimes they have no common roots in $K$, hence $P(\beta) = 0$ and $S_{P'} \subset \{\alpha\}$. The same argument works for $S_{Q'}$ and we end up showing that $S_{P'}$ and $S_{Q'}$ are two disjoint subsets of the singleton $\{\alpha\}$.
Suppose that $S_{P'} = \emptyset$ and $S_{Q'} = \{\alpha\}$. Then $n = 1 $ and $m > 1$. However, the equality $Q^2 = P^3$ implies that $3 = 2m$ which is clearly a contradiction because $m \in \mathbb{Z}$. If we suppose that $S_{Q'} = \emptyset$ and $S_{P'} = \{\alpha\}$, the same argument works and yields a similar contradiction.
If $char(K) = 3$, the equality $2P(t)Q'(t) = 3 Q(t)P'(t)$ implies that $Q'(t) = 0$. Since $G(t)P'(t) + H(t)Q'(t) = 1$ we obtain that $P'(t) = 1$ which implies $n = 1$. However, the equality $Q^2 = P^3$ implies that $3 = 2m$ which is clearly a contradiction because $m \in \mathbb{Z}$. If $char(K) = 2$ the same argument used above works and yields a similar contradiction.
This is really long and does a lot of stuff which is good to know but not really relevant to solving this problem in the intended way. You can get rid of lots of text by just jumping to the computation of $\Omega_{A/k}$: it's $A\langle dx,dy\rangle/(2ydy-3x^2dx)$, which cannot be generated by 1 element since $\Omega_{A/k}\otimes_A A/(x,y) \cong k^2$ as modules. On the other hand, $\Omega_{B/k}\cong k[t]\langle dt\rangle$ which is clearly generated by one element. Since a $k$-isomorphism $A\to B$ would give an isomorphism $\Omega_{A/k}\to\Omega_{B/k}$, we see the result.