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.
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 $xy(x+y)$.