Any triple ordinary points on an affine variety is analytically isomorphic. It is Exercise I.5.14 of Hartshorne. Let $P,Q \in V \subseteq A^{2}$ are two points such that $V$ is a curve generated by a single irreducible polynomial $f$, and by translation, if $P=(0,0)$ then $f=(ax+by)(cx+dy)(ex+gy) + f_{4} + \cdots$ and if we translate the origin to $Q$, then $f$ moved to $f'=(a'x+b'y)(c'x+d'y)(e'x+g'y) + f_{4}' + \cdots$ where each linear factors are all distinct. I want to show that those two points are analytically isomorphic, in other words, the completion of two local rings $O_{P}$ and $O_{Q}$ are isomorphic.
And in this case, $O_{P} = k[[x,y]]/(f)$ and $O_{Q} = k[[x,y]]/(f')$. From the linear algebra,  for any number $a,b,c, c'$, there exists a $2\times 2$ matrix $A$ and a scalar $M$ such that
$$A \begin{pmatrix} 1 & 1 & 1 \\ a & b & c \end{pmatrix} = M \begin{pmatrix} 1 & 1 & 1 \\ a & b & c' \end{pmatrix}.$$
Thus we may use this fact to assure that there is an automorphism sending $f'$ to $g = g_{3}+g_{4} +\cdots$ where $g_{3}=cf_{3}$ for some scalar $c$, i.e., degree 3 parts of $g$ is the same as $f_{3}$ up to scalar multiplication. This shows that we may assume degree 3 parts of $f'$ and that of $f$ shares the same linear factors.
Now, I do not know what should I do to show the ring isomorphism, since $f_{3}=g_{3}$ does not implies $k[[x,y]]/(f)\cong k[[x,y]]/(g)$ directly. One of the trial is using property of $k$-algebra homomorphism in this post, but I do not have an idea at this point to use that property.
Any hints or comments will be appreciated.
 A: Let's solve part (b) first in the interests of explaining everything (and also because the main idea of the solution exactly applies to (c) too - the slogan is "you can solve this problem term-by-term"). Recall part (b) is the statement that if $f=f_r+f_{r+1}+\cdots \in k[[x,y]]$, and $f_r$ factors as $g_sh_t$ where $f_r,g_s,h_t$ are homogeneous of degrees $r,s,t$ respectively with $g_s,h_t$ coprime, then there are formal power series $g=g_s+g_{s+1}+\cdots$ and $h=h_t+h_{t+1}+\cdots$ so that $f=gh$.
First, observe that $$gh= g_sh_t+(g_sh_{t+1}+g_{s+1}h_t)+\cdots+(\sum_{i=0}^{n} g_{s+i}h_{t+n-i})+\cdots$$ so being able to write $f=gh$ is equivalent to simultaneously solving the system of equations $f_{r+n}=\sum_{i=0}^{n} g_{s+i}h_{t+n-i}$ for all $n\geq 0$.
We proceed inductively: clearly the assumption that $f_r=g_sh_t$ means we have a solution for $n=0$.
Now assuming we've found a solution for all $n$ up to some fixed $n_0$, this means we need to solve $f_{r+n}-\sum_{i=1}^{n_0} g_{s+i}h_{t+n-i}=g_sh_{t+n_0+1}+g_{s+n_0+1}h_t$ for $g_{s+n_0+1}$ and $h_{t+n_0+1}$, and the left hand side of this equation is completely determined by our previous choices.
Let $P_d$ denote the vector space of homogeneous polynomials in two variables of degree $d$.
To show that we can always solve this equation for $g_{s+n_0+1}$ and $h_{t+n_0+1}$, we'll show that the map $P_{t+n}\times P_{s+n} \to P_{s+t+n}$ given by $(a,b)\mapsto ag_s+bh_t$ is surjective assuming $g_s$ and $h_t$ are relatively prime.
I claim it is enough to prove surjectivity for $n=0$.
Any standard basis monomial $M$ in $P_{s+t+n}$ can be written as $x^iy^j$ times some standard basis monomial $M'$ in $P_{s+t}$.
If we can find $p,q$ so that $pg_s+qh_t=M'$, then $(x^iy^jp)g_s+(x^iy^jq)h_t=M$, and this shows that every standard basis monomial in $P_{s+t+n}$ is in the image of our map, so it is surjective.
To prove surjectivity when $n=0$, consider the matrix of our map in the standard monomial basis. This is exactly the Sylvester matrix associated to the homogeneous resultant of $g_s$ and $h_t$.
But the homogeneous resultant of $g_s$ and $h_t$ is nonzero iff they are coprime, so our map is surjective since it's matrix has nonvanishing determinant.

Now on to the ordinary triple point.
Let $f=f_1+f_2+f_3+\cdots$ be a polynomial which gives an ordinary triple point at the origin.
Supposing $f_3=L_1L_2L_3$ for mutually independent linear terms, we get that $f=(L_1+\cdots)(L_2+\cdots)(L_3+\cdots)$ factors by part (b), and the automorphism given by $x\mapsto L_1+\cdots,y\mapsto L_2+\cdots$ gives that $k[[x,y]]/(f)\cong k[[x,y]]/(xy(ax+by)+g)$ where $g$ is a power series with no terms of order less than four and $a,b\neq 0$.
Up to the automorphism sending $x\mapsto bx$ and $y\mapsto ay$, we may assume that $a=b=1$, and now our goal is to eliminate $p$. Finding an automorphism of $k[[x,y]]$ which does this is equivalent to solving a collection of linear systems: the degree-$n$ portion of $xy(x+y)+g$ after substituting in $x\mapsto x+\sum_{r>1} p_r(x,y)$ and $y\mapsto y+\sum_{r>1} q_r(x,y)$ where $p_r,q_r$ are homogeneous of degree $r$ can be written as a linear combination of products of $p_i$ and $q_j$ for $i,j<n$ plus $p_n(y^2+2xy)+q_n(x^2+2xy)$.
The map from $P_n\times P_n\to P_{n+2}$ which gives the contribution of the terms $p_n,q_n$ to the degree-$n+2$ homogeneous part of our power series after substitution is $(p_n,q_n)\mapsto p_n(y^2+2xy)+q_n(x^2+2xy)$ which by the same argument as in part (b) can be seen to be surjective because $y^2+2xy=y(y+2x)$ and $x^2+2xy=x(x+2y)$ are coprime.
So we can always solve for $p_n,q_n$ to eliminate the higher-order terms, and any ordinary triple point is analytically isomorphic to the one given by $xy(x+y)$.
