How can I give an explicit presentation of the ring of differential operators over a smooth variety? I'm trying to find a presentation of the ring of differential operators for a smooth affine and smooth projective variety. For example, consider the varieties
\begin{align*}
X = \textbf{Spec}\left( R = \frac{\mathbb{C}[x,y]}{x^2 + y^2 - 1}\right) && Y = \textbf{Proj}\left( S =  \frac{\mathbb{C}[x,y,z]}{x^4 + y^4 + z^4}\right)
\end{align*}
In the first case, I can find the module of vector fields by computing the kahler differentials, and then computing the dual
$$
T_X = \text{Hom}_R\left(\Omega_{R/\mathbb{C}},R\right) = \text{Hom}_R\left(\frac{Rdx\oplus Rdy}{xdx + ydy},R\right) = \frac{R\partial_x\oplus R\partial_y}{x\partial_x + y\partial_y}
$$
and in the second case I can dualize the conormal sequence on Macaulay2.
How can I find a presentation for $D_X$ and $D_Y$? Originally I guessed that $D_X$ could be presented as
$$
\frac{R[\partial_x,\partial_y]}{(x\partial_x + y\partial_y)}
$$
but the operator on the bottom does not preserve the ideal.
 A: I think there are two mistakes made by the OP which I will correct below.

In the ring of differential operators $D$, the element $x$ actually refers to the operator of multiplication by $x$. This does not commute with $\partial_x$, instead $\partial_x(xt) = t + x\partial_xt$ or $[\partial_x,x] = 1$. Hence you cannot do $R[\partial_x,\partial_y]$, which would be a commutative ring. For instance, $$D(\Bbb C[x_i]) = \Bbb C\langle x_i,\partial_i\rangle/([x_i,x_j],[x_i,\partial_j] - \delta_{ij}\rangle.$$
Now you need to figure out the extra relations corresponding to $x^2 + y^2 - 1$ on $X$.

As you mentioned, your line $$Hom(Rdx\oplus Rdy/(2x\,dx + 2y\,dy), R) \overset{!}{=} R\partial_x\oplus R\partial_y/(2x\partial_x + 2y\partial_y)$$ is not correct and you cannot "just switch" from tangent vectors to cotangent vectors.
So how do you compute it? What you need is something such which evaluates to 0 on $2xdx + 2ydy$. So, the solution is $y\partial_x - x\partial_y$. So $D(X) = A/(x^2+y^2-1)$ where $A$ is the subring:  $$A = \mathbb C\langle x,y,y\partial_x - x\partial_y\rangle \subset D(\mathbb C[x,y])$$
You can get a presentation by combining this with the presentation above (or see the comments below this answer).

Note that this method works because $X$ is smooth. When $X$ is not smooth, you can't simply look at the ring generated by $\mathcal O_X$ and $\text{Der}(\mathcal O_X,\mathcal O_X)$, as is discussed for instance here https://cornellmath.wordpress.com/2007/09/09/d-module-basics-ii/
