Two (other) rings...are they isomorphic? Consider the local rings
$$R = \mathbb{C}[[x,y,z,w]]/\langle xyz+xyw+xzw+yzw\rangle$$
and
$$S = \mathbb{C}[[x,y,z,w]]/\langle xyz+xyw+xzw+yzw+xyzw\rangle.$$
Is $R$ isomorphic to $S$?
Some context:  I am trying to understand formal neighborhoods of points on certain varieties.  I expect one answer, and I'm getting a different answer.  This is the first nontrivial case where the answer that I get does not obviously agree with the answer that I expect.
Some history:  In a previous post (Two rings...are they isomorphic?), I asked a version of this question with one fewer variable, and Bjorn Poonen pointed out that the rings are isomorphic because there is only one kind of rational double point.  I think this was essentially an accident, hence the new post.
 A: This is very similar to Vladimir Dotsenko's approach.
The natural approach (at least, the one both he and I took) is to make a sequence of changes of variable, of the form
$$\begin{array}{rcl}
w_{n} &=& w_{n+1}+a_{n+1}(w_{n+1},x_{n+1},y_{n+1},z_{n+1}) \\
x_{n} &=& x_{n+1}+b_{n+1}(w_{n+1},x_{n+1},y_{n+1},z_{n+1}) \\
y_{n} &=& y_{n+1}+c_{n+1}(w_{n+1},x_{n+1},y_{n+1},z_{n+1}) \\
z_{n} &=& z_{n+1}+d_{n+1}(w_{n+1},x_{n+1},y_{n+1},z_{n+1}) \\
\end{array}$$
where $(a_n, b_n, c_n, d_n)$ have degree $n$ so that the polynomial $w_1 x_1 y_1 + w_1 x_1 z_1 + w_1 y_1 z_1 + x_1 y_1 z_1 + w_1 x_1 y_1 z_1$ becomes $w_n x_n y_n + w_n x_n z_n + w_n y_n z_n + x_n y_n z_n + \Delta$ for $\Delta$ of higher and higher degree. 
Thus, the key computation is the following:
If we have $wxy+wxz+wyz+xyz+\Delta$, with $\Delta$ in degrees $\geq n$, when can we make a change of variables which eliminates the degree $n$ part of $\Delta$? The answer is the following: if and only if $\Delta$ has no $w^n$, $x^n$, $y^n$ or $z^n$ term.
Proof: Let $\Delta_n$ be the degree $n$ part of $\Delta$. We can eliminate it if and only if $\Delta_n$ is of the form 
$$(wx+wy+xy) d + (wx+wz+xz) c + (wy+wz+yz) b+ (wx+wy+xy) a$$
for $a$, $b$, $c$, $d$ of degree $n-2$. In other words, if and only if $\Delta_n$ is in the ideal 
$$I:= \langle wx+wy+xy, wx+wz+xz, wy+wz+yz, wx+wy+xy \rangle.$$
It is obvious that $I$ is contained in 
$$J: = \langle wx, wy, wz, xy, xz, wz \rangle.$$
I verified by explicit linear algebra that the degree $3$ piece of $I$ has dimension $31$, as does the degree $3$ piece of $J$. So $I$ and $J$ are equal in degree $3$, and hence in all higher degrees. The degree $n$ part of $J$ is precisely the polynomials with no $w^n$, $x^n$, $y^n$ or $z^n$ term.
Thus, to win, we must show that we can keep clearing away the lowest remaining terms without creating $w^n$, $x^n$, $y^n$ or $z^n$ in higher degrees. It isn't clear to me whether or not this is possible.

Further thoughts: the germ $wxy+wxz+wyz+xyz$ is singular along $x=y=z=0$ and the permutations thereof, and likewise for $WXY+WXZ+WYZ+XYX+WXYZ$.  It seems to me that this should imply that the change of variables should take $x=y=z=0$ to $X=Y=Z=0$, and the same for the other four combinations of coordinates. This should mean that $W-w$ is in the ideal generated by $\langle w, xy, xz, yz \rangle$. Imposing these restrictions on $a$, $b$, $c$, $d$ gives a new ideal $I'$, which is only $19$ dimensional in degree $4$ (so much less than $J$.) But $wxyz$ is in $I'$, so we can still make the first change of variable.
A: This is a bit too long for a comment. 
Let me denote by $L_k=\sum_i x_i^{k+1}\frac{\partial\phantom{x_i}}{\partial x_i}$ the operators via which vector fields on a line act on polynomials in several variables (via the diagonal embedding). I noticed that if you take polynomials in $n$ variables, then the elementary symmetric function $e_n$ can be obtained as a combination of $L_1(e_{n-1})$ and $(x_1+\cdots+x_n)L_0(e_{n-1})$. Indeed, $L_1(e_{n-1})$ is the monomial symmetric function $m_{2,1^{n-1}}$, and $(x_1+\cdots+x_n)L_0(e_{n-1})=(x_1+\cdots+x_n)(n-1)e_{n-1}$ is $(n-1)(m_{2,1^{n-1}}+ne_n)$. Therefore, $e_n=\frac{1}{n(n-1)}((x_1+\cdots+x_n)L_0-(n-1)L_1)(e_{n-1})$, so $e_n$ can be obtained as the action of a vector field (infinitesimal diffeomorphisms) on $e_{n-1}$. 
This means that in the question as stated you can at least use the action of diffeomorphisms to eliminate $x_1x_2x_3x_4$ at the expense of arising higher terms (I intentionally performed the computation for any $n$ since you want it to be generalizable). 
Now specifically for $n=4$: for the next step, we have some terms of degree $5$ to kill. By construction, these terms are symmetric polynomials so they all live in a vector space of dimension $6$ (of symmetric polynomials in $4$ variables of degree $5$). At the same time, if we take "obvious" infinitesimal diffeomorphisms, that is vector fields which are combinations of $L_i$'s with coefficients being symmetric polynomials, which increase degree by $2$, then they form a $7$-dimensional space spanned by $L_2$, $e_1L_1$, $e_2L_0$, $e_1^2L_0$, $e_3L_{-1}$, $e_1e_2L_{-1}$, $e_1^3L_{-1}$. I think it is 7-dimensional, but I did not check very carefully if these are linearly independent. I also did not check carefully what happens when we apply this to $e_3$, but my feeling is that the only thing which is not hit by the action of these is scalar multiples of $\sum_i x_i^5$, and these will not appear in the infinitesimal action killing $x_1x_2x_3x_4$ either. 
My impression is that this then can propagate: at each step the only thing that is not hit by appropriate infinitesimal diffeomorphisms consists of scalar multiples of $\sum_i x_i^k$, and these never have to be eliminated. 
I am sorry for not filling in the details, but I have to finish a lot of stuff today (this lovely question already distracted me more than it should have!)
A: Consider the map $\varphi:S\to R$ given by
$$\varphi(x) = \frac{4x}{4-x},$$
and similarly for $y$, $z$, and $w$.  One needs to check that this is well-defined;
indeed, $\varphi$ maps $xyz+xyw+xzw+yzw+xyzw$ to 
$$\frac{4^4(xyz+xyw+xzw+yzw)}{(4-x)(4-y)(4-z)(4-w)}.$$
Since $\varphi$ obviously induces an isomorphism on the associated graded of the filtration by powers of the maximal ideal, it is itself an isomorphism.
More explicitly, the inverse homomorphism $\psi:R\to S$ is given by
$$\psi(x) = \frac{4x}{4+x},$$
and similarly for $y$, $z$, and $w$.
Note also that this generalizes to any number of variables.
Thanks David and Vladimir for your answers; that was really helpful for getting me going!
