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 ndimensional complex projective space.

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 $$X^3+Y^3+Z^33(1+\lambda^2)XYZ=0.$$ 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 $S^23XY=0$. 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 $I_6$. 


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:



A simple computation shows that the equation $$ u(x^3+y^3+z^3)3vxyz=0 $$ defines a nonsingular 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 $$(\xi,\eta)\mapsto \zeta=\xi\eta.$$
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 $$\zeta^2=\xi\eta.$$
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.



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 twotoone eveywhere except at $s=0$ and at $t=0$ where it is onetoone. 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 $\mathbb{P}^1$. 

