Does formal smoothness of $R \to R[x,y]/(f)$ imply $(f,f_x,f_y)=R[x,y]$? I came across this question about formal smoothness implying smoothness in a specific example. That question was old and unanswered, so I'm taking the opportunity to ask it again. 
I tried to use the hint in Mariano's comment in the above-linked question to show that $(f_x,f_y)=(1)$ in the ring $R[x,y,s,t]/(f(x+s,y+t),s^2,st,t^2)$, but I essentially got nowhere. 
I also came across some other similar questions here, such as this question, but none of them have concrete answers. I looked at theorem 30.3 (p.233) of Matsumura's Commutative Ring Theory and also this link, but I just cannot see how to adapt those local arguments to this example.
So is it possible to show the Jacobian criterion holds when $R$ is this general explicitly (that is, without having to appeal to the conormal sequence and differentials)? Is it easier to show if $R$ is a field? 
 A: Let $A = R[x,y]/(f)$ and $A' = A[\varepsilon]/(f-\varepsilon)$ and $\varepsilon^2 = 0$. Then there is an exact sequence $$\varepsilon A \to A' \to A$$ since $(\varepsilon) \cong A$ as $A$-modules. This algebra $A'$ is the same as $R[x,y]/(f)^2$, the first infinitesimal neighborhood of the embedding $\text{Spec}\, A \to \mathbb A_R^2$, I just like the epsilons.
Now suppose you have a splitting $s: A \to A'$. Then $s$ sends
\begin{align*}
x&\mapsto x + \varepsilon p_1\\
y&\mapsto y + \varepsilon p_2
\end{align*}
for some $p_1,p_2 \in A$. It's a homomorphism, so it also sends 
\begin{align*}
f&\mapsto f(x+\varepsilon p_1, y+\varepsilon p_2)\\
&= f + \frac{\partial f}{\partial x}\varepsilon p_1 + \frac{\partial f}{\partial y} \varepsilon p_2\\
&=\varepsilon\left(1 + \frac{\partial f}{\partial x} p_1 + \frac{\partial f}{\partial y} p_2\right)
\end{align*}
The second line is true even in $R[x,y,\varepsilon]/(\varepsilon^2)$; in the third line we have used $f = \varepsilon$ from $A'$.
But $f=0$ in $A$, so the right hand side must also be zero in $A'$, which translates to $$f\ \ \bigg| \left(1 + \frac{\partial f}{\partial x} p_1 + \frac{\partial f}{\partial y} p_2\right)$$
i.e., whenever $f$ vanishes (on $\text{ Spec } A$) then that Jacobian matrix of $f$ does not vanish (Jacobian criterion), and $1 = - p_1f_x - p_2 f_y$ mod $f$.
A generalization:
Note that the above argument is also easily generalized to $A = R[x_1,\ldots, x_n]/(f_1,\ldots, f_r)$, taking $A' = R[x_1,\ldots, x_n, \varepsilon_1,\ldots, \varepsilon_r]/(f_i - \varepsilon_i)$ and $\varepsilon_i\varepsilon_j =0$ (the first infinitesimal neighborhood again).
One assumes sections $x_k \mapsto x_k + \sum_j\varepsilon_jp_{kj}$, then considers where the $f_i$ are sent. As above, one gets equations for each $i$:
$$f_i \mapsto \sum_j \varepsilon_j\left(\delta_{ij} + \sum_{k=1}^n \frac{\partial f}{\partial x^k}p_{kj}\right)$$
setting this to zero gives $r\times r$ equations which must vanish. Putting these into matrices, this says $$I + (Df)P = 0 \text{ mod } f_1,\ldots, f_r$$ for some $n\times r$ matrix $P$, where $I$ is the $r\times r$ matrix of rank $\text{min}(r,n)$, consisting of $\delta_{ij}$ for $1\leq i,j \leq \text{min}(r,n)$ and zeros. In other words, on $\text{Spec} A$ the Jacobian $Df$ has full rank.
This shows that formal smoothness implies smoothness, for varieties.
