How to resolve the singularity of $xy+z^4=0$? This singularity can not be resolved by one time blow-up.  I don't know how to blow up the singularity of the "variety" obtained by the first blow-up, in other words, I am confused with how to do the successive blow-ups. Does it have a explicit form in affine coordinates? And how to find the exceptional divisor in this case? Please show the process in detail.
Here is my attempt:
Let $(x,y,z;p_1,p_2,p_3)$ be the coordinates of $\mathbb{C}^3\times\mathbb{P}^2$ , after blow-up at $(0,0,0)$ we have three pieces:
$p_1=1, x^2+p_2p_3=0$
$p_2=1, p_3+p_1^4y^2=0$
$p_3=1, p_2+p_1^4z^2=0$
And from the Jacobian matrix, the first piece still has a singularity. We need to blow-up again. But it is not the subvariety of $\mathbb{C}^3$ anymore, we can not use the technique again. My problem is how to do in the next.
P.S. There is a similar problem in the Hartshorne's book(the exercise 5.8 in Chapter V.)
P.S.S. I am unfamiliar with the intersection theory, so I want to work this out in the rudimentary way...
 A: I don't know if you are still interested, here is a more detailed answer. By the change of variable $a = x+iy, b=x-iy, c = z$ your singularity becomes $a^2 + b^2 + c^4 = 0$, and indeed it is a $A_3$ singularity, so I will assume the equation is $x^2 + y^2 + z^4 = 0$. 
For the blow-up, I will take affine coordinates, so performing several blow-up is more easy.
The first chart will have coordinates $x, y_1 := y/x, z_1 = z/x$ (which corresponds to the usual charts on $\Bbb P^2$), so $xy_1 = y$ and $xz_1 = z$. Plugging these relations we get the equation $x^2(1 + y_1^2 + x^2z_1^4) = 0$. This is a reducible surface, the $x^2 = 0$ term correspond to the exceptional divisor $E$ of the blow-up of $\Bbb C^3$, the $1 + y_1^2 + x^2z_1^4 = 0$ term is the strict transform $X'$ of your surface which is smooth so we don't need to worry about this chart, similarly for the chart with coordinates $x/y,z/y,y$. 
Now let's look at the last chart, i.e the coordinates are $x_2,y_2,z$ where $zx_2 = x$ and $zy_2 = y$. The equation becomes $z^2(x_2^2 + y_2^2 + z^2) = 0$. The strict transform is singular (it is the $A_1$ singularity). The exceptional divisor $E_1$ is given by $z = 0, x_2^2 + y_2^2 = 0$ i.e we obtain two lines, intersecting at $(0,0,0)$ which is the singular point. Spoiler : after one further blow-up, they will be separated and we will get the $A_3$ configuration as expected.
Let's check it, and perform a second blow-up. I will check everything in one chart and you can check for yourself that nothing more happens in the two other charts.
After one blow-up, our surface equation was given by $x_2^2 + y_2^2 + z^2 = 0$ , and the exceptional divisor $E_1$ was given by $z = 0, x_2^2 + y_2^2 = 0$, i.e $z = 0, x_2 = \pm i y_2$. 
We will take the following coordinate for the second blow-up : $x_3,y_2,z_3$ with $y_2x_3 = x_2, z_3y_2 = z$. The equation becomes $y_2^2(x_3^2 + 1 + z_3^2) = 0$. As usual, $y_2^2 = 0$ corresponds to the exceptional divisor of the blow-up of $\Bbb C^3$ and $x_3^2 + 1 + z_3^2 = 0$ is the strict transform of the surface, which is smooth. The exceptional divisor $E_2$ is given by $y=0, 1 + x_3^2 + z_3^2 = 0$ i.e it's a smooth conic. Now, the strict transform of $E_1$ is given by the equation $z_3 = 0, x_3 = \pm i$. We see that $E_2$ intersects each component of $E_1$ at $(\pm i, 0, 0)$ and I claim that the two component of $E_1$ do not intersect anymore, so we get indeed the dual configuration of the Dynkin diagram $A_3$. 
A: This is a $2$-dimension cyclic quotient singularity. Here is a very nice PDF file which explains the classification of two dimensional rational double points.
Here is a sketch on how to blow up. You start by blowing up $\mathbb A^3$ in the origin. This is given by the set of equations $xs=ry,xt=rz$ and $yt=sz$, where $(r:s:t)$ are coordinates in $\mathbb P^2$. These are the $2 \times 2$ minors of the matrix 
$$
\begin{pmatrix}
x & y & z \\
r & s & t
\end{pmatrix}.
$$
So you get three affine charts. Putting $s=1$ and $r=1$, you can eliminate $y$ and $z$ and $x$ and $z$, respectively. In each case, you get a smooth variety.
In the chart $t=1$, however, you get the singular variety $rs+z^2=0$. But this is the ordinary double point, which is resolved after one more blowup. One way to resolve this, is to look at the map $r=az, s=bz$ and $z=z$. Then you get $z^2(ab+1)$. The strict transform is $ab+1=0$ (in $\mathbb A^3$ with coordinates $a,b,z$), with exceptional divisor given by $z=0$.
Hence you need two blowups to resolve the singularity. By the classification of rational double points, this is a $A_2$ singularity. 
