Proof. We can rewrite Lemma 12.3.4 as
\begin{align*}\pd_{t}y_{m}=\sums{n-1 }{i=1}\sums{N }{l=1}a^{i}_{ml}(x,y_{1},...,y_{N})\pd_{i}y_{l}+b_{m}(x,y_{1},...,y_{N})\end{align*}
We seek $y_{m}$ of the form $y_{m}=\sums{}{}c_{m}^{\al j}x^{\al} t^{j}$. Impose $Y(x,0)$
\begin{align*}c_{m}^{\al0}=0&&\fall \al,m\end{align*}
So we must determine $c_{m}^{\al j}$ for $j>0$. Substitute $y_{m}$ into the PDE, by Lemma 12.3.7(2) the RHS is
\begin{align*}\sums{}{\al,j}P_{m}^{\al j}\parens{(c_k^\be l)_{k\leq j},\text{coeffs of $A_{i},B$}}x^{\al} t^{j}\end{align*}
with non-negative coefficients. The LHS is
\begin{align*}\sums{}{\al,j}(j+1)c_{m}^{\al j+1}x^{\al} t^{j}\end{align*}
So equating coefficients of $x^{\al+j}$
\begin{align*}c_{m}^{\al j+1}=\frac{1}{j+1}P_{m}^{\al j}(c,\text{coeff of $A_{i},B$})\end{align*}
Inductively we have polynomials of non-negative coefficients $Q_{m}^{\al j}$
\begin{align*}c_{m}^{\al j}=Q_{m}^{\al j}(\text{coeff of $A_{i},B$})\end{align*}
Next we construct a system
\begin{align*}\begin{cases}\pd_{t}\til{Y}=\sums{n-1}{i=1}\til{A_i}(x,\til{Y})\pd_{i} \til{Y}+\til{B}(x,\til{Y})\\ \til(Y)(x,0)=c\end{cases}\end{align*}
such that $\til{A_i},\til{B}$ are $C^{\om}$, the system has a $C^{\om}$ solution near $0\in \R^{n-1}\ti\R$ and Taylor series of $\til{A_i},\til{B}$ majorizes those of $A_{i},B$. By Lemma 12.3.7(3,5), if $M>0$ is sufficiently large and $r>0$ is sufficiently small, the series for $A_{i},B$ can be majorized by (geometric) series for
\begin{align*}\frac{Mr}{r-\sums{n-1}{i=1}x_{i}-\sums{N }{j=1}y_{j}}\end{align*}
Consider now the Cauchy problem
\begin{align*}\begin{cases}\pd_{t}y_{m}=\frac{Mr}{r-\sums{n-1}{i=1}x_i-\sums{N }{j=1}y_j}\parens{\sums{n-1}{i=1}\sums{N}{j=1}\pd_iy_j+1}\\ y_{m}(x,0)=0\end{cases}\end{align*}
It remains to show this has a $C^{\om}$ solution near $0$. By utilizing symmetry we claim $u$ as follows is such a solution. Let $y_{j}\define u(x_{1}+\cds+x_{n-1},t)_{1\leq j\leq N}$ where $u=u(s,t)$ solves
\begin{align*}\begin{cases}\pd_{t}u=\frac{Mr}{r-s-Nu}\parens{N(n-1)\pd_su+1}\\ u(s,0)=0\end{cases}\end{align*}
We now solve for $u$ by method of characteristic.
\begin{align*}\der{t}{\tau}=r-s-Nu&&\der{s}{\tau}=-MrN(n-1)&&\der{u}{\tau}=Mr\\ t(0)=0&&s(0)=\s\text{ parametrized along initial curve}&&u(0)=0\end{align*}
where $\tau$ is the flow parameter, then $\der{^2t}{\tau^2}$ is constant. After some simple calculations
\begin{align*}t&=\frac{1}{2}MrN(n-2)\tau^{2}+(r-\s)\tau\\ u&=Mr\tau\\ s(\tau)&=-MrN(n-1)\tau+\s\\ u(s,t)&\define u(\tau(s,t),\s(s,t))\\&=Mr\tau(s,t)\\&=\frac{r-s-\sqrt{(r-s)^{2}-2MrNt}}{Mn}\in C^{\OM}\text{ near 0}\\ u(s,0)=0\end{align*}
$\square$
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