As Alex Woo says, this is a toric example, and hence can be solved with toric methods. Your variety is $\mathrm{Spec} \ \mathbb{C}[S]$ where $S$ is the semigroup ring generated by $(1,0,0,1)$, $(0,1,0,1)$, $(0,0,1,1)$, $(0,0,0,1)$ and $(1,1,1,2)$. (These correspond to the variables $x_1$, $x_2$, $x_3$, $y_1$ and $y_2$ respectively.) This is a saturated semigroup, so toric methods apply with no subtleties. The dual cone is generated by $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$, $(-1,-1,0,1)$, $(-1,0,-1,1)$, $(0,-1,-1,1)$. Toric resolutions of this singularity correspond to triangulations of this cone.

Notice that the six generators of the dual cone all lie in the plane $w+x+y+3z=1$. In this plane, they form a triangular prism. We can draw our pictures in three coordinates by discarding the final coordinate. However, we need to recognize that, if we do this, a lattice point means a point $(x,y,z)$ such that $x+y+z \equiv 1 \mod 3$. (This is the point I screwed up earlier.)

So we want to understand triangulations of the prism with vertices $(1,0,0)$, $(0,1,0)$, $(0,0,1)$, $(0,-1,-1)$, $(-1,0,-1)$, $(-1,-1,0)$. Now, suppose that our subdivision has a face $d$ which is contained in a face of the triangular prism of dimension $e$. The cones on these faces have dimension $d+1$ and $e+1$; the corresponding torus orbits have dimension $3-d$ and $3-e$, so the fibers here have dimension $e-d$. So smallness means that $2(e-d) < e+1$. For $e=0$, $1$, $2$, $3$ this gives $d \geq 0$, $1$, $1$ and $2$, respectively. In other words, we must add no new vertices to the triangular prism, and we may only add new edges within $2$-faces.

Fortunately, the standard triangulation of the triangular prism has this property.
There are three tetrahedra:
$$\mathrm{Hull} {\Large (} (1,0,0), \ (0,1,0), \ (0,0,1), \ (0,-1,-1) {\Large )}$$
$$\mathrm{Hull} {\Large (} (0,1,0), \ (0,0,1), \ (0,-1,-1), \ (-1,0,-1) {\Large )}$$
$$\mathrm{Hull} {\Large (} (0,0,1), \ (0,-1,-1), \ (-1,0,-1), \ (-1,-1,0) {\Large )}$$
In my previous update, I worried that these are not unimodular, because the "lattice point" $(0,0,0)$ lay on the $2$-faces of some of them. However, that is actually not a lattice point. (It corresponds to $(0,0,0,1/3)$ back in $4$-space.) Sorry about the confusion.