Derivations of the algebra $K[x,y]/(y^2-x^3)$ Let $K$ be a field of characteristic zero, and let $S=K[x,y]/(y^2-x^3)$. It is easy to see that $S\cong R:=K[t^2,t^3]$ via the isomorphism induced by the ring homomorphism $K[x,y]\to R$ given by $x\mapsto t^2, y\mapsto t^2$. It is known that for any $K$-algebra $A$ and any ideal $J$,
$$
\operatorname{Der}_K(A/J) \cong \operatorname{Der}_J(A)/J\operatorname{Der}_K(A),
$$
where $\operatorname{Der}_K(A)$ is the space of $K$-linear derivations of $A$, and
$$
\operatorname{Der}_J(A) = \{D\in \operatorname{Der}_K(A) \; | \; D(J)\subseteq J\}.
$$
In Exercise 3.6 of Chapter 3 in the book A primer on Algebraic $D$-modules by Coutinho, I'm asked to show that the derivations
$$
D_1 = 2y\partial_x + 3x^2\partial_y \quad \text{and} \quad D_2 = 3y\partial_y - 2x\partial_x
$$
generate $\operatorname{Der}_K(S)$ as $S$-module. Then I have to conclude that $\operatorname{Der}_K(R)$ is generated as an $R$-module by $t\partial_t$ and $t^2\partial_t$.
I have two issues here: First, it seems to me that $D_2$ must be $3y\partial_y + 2x\partial_x$, as the application of the chain rule gives
$$
t\partial_t f = t \frac{\partial f}{\partial x} \frac{dx}{dt} + t \frac{\partial f}{\partial y} \frac{dy}{dt} = 2t^2 \frac{\partial f}{\partial x} + 3t^3 \frac{\partial f}{\partial y} = (2x\partial_x + 3y\partial_y)f
$$
and
$$
t^2\partial_t f = t^2 \frac{\partial f}{\partial x} \frac{dx}{dt} + t^2 \frac{\partial f}{\partial y} \frac{dy}{dt} = 2t^3 \frac{\partial f}{\partial x} + 3t^4 \frac{\partial f}{\partial y} = (2y\partial_x + 3x^2\partial_y)f = D_1f.
$$
Is it a typo in the book or am I misunderstanding something?
The second one is about the proof that $D_1$ and $D_2$ generates $\operatorname{Der}_K(S)$ as $S$-module. First of all, an element of $\operatorname{Der_J(K[x,y])}$, where $J=(y^2-x^3)$ has the form $d=f_1 \partial_x + f_2\partial_y$, and the condition $d(J)\subseteq J$ means that
$$
-3x^2f_1 + 2yf_1 \in J,
$$
and the condition that that $D_1$ and $D_2$ generates $\operatorname{Der}_K(S)$ as $S$-module means that there are polynomials $g,h\in K[x,y]$ such that
$$
f_1 \partial_x + f_2\partial_y - gD_1 -hD_2 \in J\operatorname{Der}_K(K[x,y]),
$$
but I can't manage to show that such $g$ and $h$ would exist.
Working a little bit this attempt, I need to find polynomials $u,v\in K[x,y]$ such that
$$
f_1 \partial_x + f_2\partial_y - gD_1 -hD_2 = (y^2-x^3)(u\partial_x + v\partial_y).
$$
Taking $D_2 = 3y\partial_y + 2x\partial_x$ with the "right" sign, we can see that the above equation is equivalent to the following equalities:
$$
f_1 - 2yg - 2xh = (y^2-x^3u) \quad \text{and} \quad f_2 - 3x^2g - 3yh = (y^2-x^3)v,
$$
and, as $-3x^2f_1 + 2yf_2 \in J$ there exists some polynomial $w\in K[x,y]$ such that $-3x^2f_1 + 2yf_2=(y_2-x^3)w$. A little manipulation gives
$$
-6h = -3x^2v + 2yu - w
$$
and I don't know if I can do more, or even if my attempt is correct, so any help will be appreciated.
 A: In terms of the modules of Kähler differentials, we have an exact sequence of $S$-modules
$$J/J^2 \overset{\delta}{\rightarrow} S \,dx \oplus S \,dy \rightarrow \Omega^1_{S/K} \to 0.$$
(Here, the middle term is $\Omega^1_{K[x,y]/K} \otimes_{K[x,y]} S$, and $J / J^2 \simeq S$; and $\delta : y^2 - x^3 \mapsto 2y\,dy - 3x^2\,dx$.)  Taking the duals, we get an exact sequence
$$0 \to \operatorname{Der}_K(S) \to S \partial_x \oplus S \partial_y \to S.$$
Therefore, $\operatorname{Der}_K(S)$ is the kernel of the map $\varphi : S \partial_x \oplus S \partial_y \to S$ given by $\partial_x \mapsto -3x^2, \partial_y \mapsto 2y$.  (In particular, I think you're right that there's a typo in the book: $3y \partial_y + 2x \partial_x \mapsto 3y (2y) + 2x (-3x^2) = 6(y^2 - x^3) = 0$; so $3y \partial_y + 2x \partial_x$ is indeed in the kernel.)
Note that the above is essentially equivalent to your calculation based on $\operatorname{Der}_K(A/J) \simeq \operatorname{Der}_J(A) / J \operatorname{Der}_K(A)$ where $\operatorname{Der}_J(A) = \{ \partial \in \operatorname{Der}_K(A) \mid \partial J \subseteq J \}$.
Now, it is easy to see that the graph of $\varphi$, as a submodule of $S \partial_x \oplus S \partial_y \oplus S$, is generated by
\begin{align*}
e_1 & = ( 1, 0, -3x^2 ) \\
e_2 & = ( 0, 1, 2y).
\end{align*}
This is equivalent to looking at the $K[x,y]$-submodule of $K[x,y] \partial_x \oplus K[x,y] \partial_y \oplus K[x,y]$ generated by
\begin{align*}
\bar e_1 & = (1, 0, -3x^2 ) \\
\bar e_2 & = (0, 1, 2y) \\
f_1 & = (y^2 - x^3, 0, 0) \\
f_2 & = (0, y^2 - x^3, 0) \\
f_3 & = (0, 0, y^2 - x^3).
\end{align*}
From here, calculate a Gröbner basis for this submodule with respect to an elimination order in which the domain coordinates are smaller than the codomain coordinates.  That will make it easy to get a generating set for the intersection of the graph of $\varphi$ with $S \partial_x \oplus S \partial_y \oplus 0$, from which it will be easy to deduce a generating set for $\operatorname{ker}(\varphi) \simeq \operatorname{Der}_K(S)$.
