minimal resolution of singularities What is the minimal resolution of singularities of the surface
$S^2(X^3+Y^3+Z^3)-3(S^2+T^2)XYZ=0$ which is a subset of $\mathbb{P}^1\times\mathbb{P}^2$
 Please note that in this equation $[S:T]\in{\mathbb{P^1}}$ and $[X:Y:Z]\in{\mathbb {P^2}}$ and by $\mathbb{P^n}$ we mean n-dimensional complex projective space.
 A: The deleted comment was mine - I just stated that the singular locus consisted of six isolated singular points. Here is a Macaulay2 session to back up this claim:
Resolution of the surface S^2(X^3+Y^3+Z^3)-3(S^2+T^2)XYZ=0 in P2xP1:

Macaulay2, version 1.4
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases,
               PrimaryDecomposition, ReesAlgebra, TangentCone

i1 : R:=QQ[x,y,z,s,t]

o1 : PolynomialRing

i2 : I=ideal(s^2*(x^3+y^3+z^3)-3*(s^2+t^2)*x*y*z)

            3 2    3 2           2    3 2           2
o2 = ideal(x s  + y s  - 3x*y*z*s  + z s  - 3x*y*z*t )

o2 : Ideal of QQ[x, y, z, s, t]

i3 : hyper=R/I

i4 : 
     P5=QQ[v_0..v_5] -- making a map to P5 using the Segre embedding

o4 : PolynomialRing

i5 : segre=map(hyper,P5,matrix{{x*s,y*s,z*s,x*t,y*t,z*t}});

o5 : RingMap hyper <--- P5

i6 : J=ker segre

                                                    2      2                2  
o6 = ideal (v v  - v v , v v  - v v , v v  - v v , v v  + v v  - 3v v v  + v v 
             2 4    1 5   2 3    0 5   1 3    0 4   0 3    1 4     0 1 5    2 5
     --------------------------------------------------------------------------
                 3    3              3
     - 3v v v , v  + v  - 3v v v  + v  - 3v v v )
         3 4 5   0    1     0 1 2    2     0 4 5

o6 : Ideal of P5

i7 : 
     V=variety(J)  -- this is the surface in P5

o7 = V

o7 : ProjectiveVariety

i8 : dim V

o8 = 2

i9 : I=ideal singularLocus V -- this is the singular locus, it has dimension 0


o9 = ideal (v v , v v , v v , v v , v v , v v , v v , v v , v v , v v , v v ,
             4 5   3 5   2 5   1 5   0 5   3 4   2 4   1 4   0 4   2 3   1 3 
     --------------------------------------------------------------------------
            2                 2   2
     v v , v  - v v , v v  - v , v  - v v )
      0 3   1    0 2   0 1    2   0    1 2

o9 : Ideal of P5

i10 : decompose I  -- Primary decomposition


o10 = {ideal (v , v , v , v , v ), ideal (v , v , v , v , v ), ideal (v , v ,
               2   1   0   4   3           2   1   0   5   3           2   1 
      -------------------------------------------------------------------------
                                                     2           2
      v , v , v ), ideal (v  + v  + v , v , v , v , v  + v v  + v ), ideal (-
       0   5   4           0    1    2   5   4   3   1    1 2    2           
      -------------------------------------------------------------------------
      v  + v , v  - v , v , v , v )}
       1    2   0    2   5   4   3

o10 : List

i11 : W=variety(I)

o11 : ProjectiveVariety

i12 : dim W

o12 = 0

i13 : degree I -- and the sigular locus consists of 6 points, each with multiplicity 1

o13 = 6

A: A simple computation shows that the equation
$$
u(x^3+y^3+z^3)-3vxyz=0
$$
defines a non-singular surface $F\subset\mathbb P^1\times \mathbb P^2$. The projection to $\mathbb P^1$ gives an elliptic fibration $\sigma:F\to \mathbb P^1$. This has exactly two singular fibers, over $[0:1]$ and $[1:1]$, each consisting of three lines not going through a common point. The local equation for the projection at the singular points of the fibers is $$(\xi,\eta)\mapsto \zeta=\xi\eta.$$

Details (to satisfy popular demand): 
a) Near the point $[0:1]\times[0:0:1]$, let $\zeta=\dfrac uv$, $\xi=\left(\dfrac{3vz}{x^3+y^3+z^3}\right)\cdot x$, and $\eta=y$. Notice that $\dfrac{3vz}{x^3+y^3+z^3}$ is a unit near that point. Near the other singular points of the fiber over $[0:1]$ permute the variables accordingly.
b) Near the point $[1:1]\times [1:1:1]$, let $\zeta=\dfrac{v-u}u$, $\xi=\left(\dfrac{x+y+z}{3xyz}\right)\cdot(x+\omega y+\omega^2 z)$, and $\eta=x+\omega y+\omega^2 z$ where $\omega\neq 1$ is a $3$rd root of unity. In particular $1+\omega+\omega^2=0$. Notice that $x^3+y^3+z^3-3xyz=(x+y+z)(x+\omega y+\omega^2 z)(x+\omega y+\omega^2 z)$. Permute the three linear factors accordingly for the other two singular points of the fiber.

Note: Actually one can conclude the stated local condition without doing this explicit calculation. The point is this: we know that the singular fiber is three lines in the plane intersecting in three separate points. Therefore, locally each of the singularities of the fiber is defined by $\zeta=\xi\eta$. Since the nearby fibers are smooth, the family, locally, is a smoothing of a node. The versal deformation space of a node is one dimensional (it's exactly what the displayed equation claims) and hence this smoothing has to be locally isomorphic to that.  


Now consider a base change of $\sigma$ by taking square roots $\mathbb P^1\to \mathbb P^1$, $[s:t]\mapsto [s^2:s^2+t^2]$. The new surface $G=F\times_{\mathbb P^1}\mathbb P^1$ is the surface in the question. This will acquire singularities over the points where $\sigma$ was not a smooth morphism. We saw above that the local equation of the map at those points is given by 
$$(\xi,\eta)\mapsto \zeta=\xi\eta.$$
The base change replaces $\zeta$ by $\zeta^2$, so the local equation of the surface becomes $$\zeta^2=\xi\eta.$$

Details:
a) near $[0:1]$ we had above $\zeta=\dfrac uv$, so the base change makes it $\zeta=\dfrac{s^2}{s^2+t^2}=\left(\dfrac 1{1+\tau^2}\right)\rho^2$. Replace $\zeta$ with $\rho$ and $\xi$ with $\xi\cdot(1+\tau^2)$.
b) near $[1:1]$ we had $\zeta=\dfrac{v-u}u$, so the base change makes it $\zeta=\dfrac{t^2}{s^2}=\rho^2$. Replace $\zeta$ with $\rho$.

Note: Again, this can be done without the explicit computation. Any two-to-one map $\mathbb P^1\to \mathbb P^1$ is simply taking roots of the local coordinates defining the points where that map is branched. Therefore if $\zeta$ is the local equation of the branch point, then the cover replaces $\zeta$ with $\zeta^2$.


In other words, the surface has exactly $6$ singular points, each locally analytically isomorphic to the vertex of a quadratic cone, and hence blowing up these points (once) yields the minimal resolution.

Edit history:
  1) Thanks to JME for pointing out the typo in the definition of the base change map.
2) Edit 1: added the local calculation for the description of the map near the singular points.
3) Edit 2: added the theoretical argument (which in my mind actually preceded the calculation) that implies the same result as the calculation.

A: Francesco explained  beautifully the resolution. Since I had prepared  a geometric description of the resolution, I  thought I will still post it. 
The singular surface 
$$
E: S^2 (X^3+Y^3+Z^3)-3 (S^2+T^2) X Y Z=0
$$
is an hypersurface of  bidegree $(2,3)$ in $\mathbb{P}^1\times \mathbb{P}^2$. 
The rational curve $\mathbb{P}^1$ is parametrized by the projective coordinates $[S:T]$ and 
$[X:Y:Z]$ are projective coordinates of $\mathbb{P}^2$. For every point of $\mathbb{P}^1$, the equation defines a cubic in $\mathbb{P}^2$ which is in the form of Hesse pencil:
$$
H:  s (X^3+Y^3+Z^3)+ t XYZ=0, \quad [s:t]\in \mathbb{P}^1.
$$ 
Hesse pencil is famous in number theory, in cryptography and also shows up  examples of mirror symmetry in physics. 
It is related to the Hesse configuration of 9 points and 12 lines in $\mathbb{P}^2$. 
There is a nice review by Artebani and Dolgachev. 
Hesse pencil can be seen as an elliptic surface with base $\mathbb{P}^1$. It admits singular fibers of Kodaira type $I_3$ (three lines forming a triangle).
The fibration considered in the question is obtained from Hesse pencil with the following map:
$$
[s:t]\mapsto [s^2:-3(s^2+t^2)].
$$
This map is two-to-one  eveywhere except at $s=0$ and at $t=0$ where it is one-to-one. 
This is related to the $\mathbb{Z}_2$ singularities described by Francesco in his answer. 
The six singular points of the elliptic surface $E$ are the intersection points of the three lines that form the  fibers $I_3$ above  $[S:T]=[1:0]$ and $[S:T]=[0:1]$. After the resolution, the singular points are replaced by $(-2)$-curves.  The resolution describes a topological transition where two singular fibers of type  $I_3$ are replaced by fibers of Kodaira type $I_6$. The transition is realized by replacing on each $I_3$ fiber, each of the  3 intersections points of  the three lines by  a $\mathbb{P}^1$. 
A: Let us start by writing down the computation of the singular points in the chart $S=1$. 
Writing $\lambda:=T/S$, in the chart $S=1$ we can rewrite the equation of the surface as
$$X^3+Y^3+Z^3-3(1+\lambda^2)XYZ=0.$$
This is an elliptic fibration over $\mathbb{C}$ (with coordinate $\lambda$), whose fibres are the curves of the Hesse pencil of cubics in $\mathbb{P}^2$.
Taking derivatives with respect to $X, Y, Z, \lambda$ we obtain the equations:
$$X^2-(1+\lambda^2)YZ=0,$$ $$Y^2-(1+\lambda^2)XZ=0,$$ $$Z^2-(1+\lambda^2)XY=0,$$ $$\lambda XYZ=0.$$
The only possibility is $\lambda=0$, so the singularities are the three points
$$[1:1:1], \;  [1: a :a^2], \; [1:a^2:a], \quad a:=e^{2 \pi i /3}$$
in the fibre over $\lambda=0$. In fact, the fibre over $\lambda=0$ degenerates as the union of three distinct lines, which form a triangle whose vertices are the three points above.
An easy local computation shows that all these points are of type $A_1$, so the minimal resolution for each of them is given by a $(-2)$-curve. In other words, the fibre of the resolved surface in $\lambda=0$, i.e over $[S:T]=[1:0]$, is of type $I_6$ according to Kodaira classification.   
Now let us consider the chart $T=1$. The equation of the surface becomes
$$S^2(X^3+Y^3+Z^3)-3(S^2+1)XYZ=0.$$
We are interested only on the singularities lying over $S=0$, and a straightforward computation gives the three points 
$$[1:0:0], \; [0:1:0], \; [0:0:1].$$
In fact, the fibre over $[S:T]=[0:1]$ degenerates to $XYZ=0$, i.e. the union of the three coordinate lines.
In the chart $Z=1$ the equation becomes
$$S^2(X^3+Y^3+1)-3(S^2+1)XY=0,$$
so the tangent cone in $(X,\,Y)=(0,\,0)$ is the irreducible quadric $S^2-3XY=0$. In the other charts the situation is the same, so again we have three points of type $A_1$.
Summing up, the surface has three points of type $A_1$ over $[S:T]=[1:0]$, three points of type  $A_1$ over $[S:T]=[0:1]$ and no other singularities. 
The minimal resolution is an elliptic fibration over $\mathbb{P}^1$ with two reducible fibres of type  $I_6$.
