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.

nottrue in general that the analytic (or formal) type of the singularity is the same of the type of the tangent cone. For instance, take $$\mathbb{C}[[x, y, z, w]]/(x^3+y^3+z^3+w^3), \quad \mathbb{C}[[x, y, z, w]]/(x^3+y^3+z^3+w^3+xyzw).$$ Then these two germs arenotisomorphic: in fact they have the same Milnor number ($16$) but different Tjurina number ($16$ and $15$, respectively). $\endgroup$ – Francesco Polizzi Nov 12 '14 at 15:02