implies equation  and it remains to prove .

Suppose $1\le\alpha\le N$ and $\beta=N+1$. Since $[xw]=v_{N+1}[w]$, we have $(c^{\mathrm{ext}}_v)^{N+1}_{k,N+1}=v_{N+1}^{k-1}$ for $1\le k\le N$, and, therefore, $$(c^{\mathrm{ext}})^{N+1}_{\alpha,N+1}=\sum_{k=1}^N\frac{\d v_k}{\d t^\alpha}(c^{\mathrm{ext}}_v)^{N+1}_{k,N+1}=\sum_{k=1}^N\frac{\d v_k}{\d t^\alpha}v_{N+1}^{k-1}\stackrel{\text{eq.\eqref{eq:FoAN and the flat coordinates}}}{=}\frac{\d^2 F^o_{A_N}}{\d t^\alpha\d t^{N+1}}.$$

Suppose $\alpha=\beta=N+1$. Then we compute $$(c^{\mathrm{ext}})^{N+1}_{N+1,N+1}=(c^{\mathrm{ext}}_v)^{N+1}_{N+1,N+1}\stackrel{\text{eq.\eqref{eq:w2 for AN}}}{=}v_{N+1}^N+\sum_{k=2}^N(k-1)v_k v_{N+1}^{k-2}\stackrel{\text{eq.\eqref{eq:FoAN and the flat coordinates}}}{=}\frac{\d^2 F^o_{A_N}}{\d (t^{N+1})^2}.$$

Finally, if $1\le\alpha,\beta\le N$, then from the associativity of the algebra ${{\widehat{\mathcal{A}}}}^{{\mathrm{ext}}}_{A_N}$ we get $$(c^{{\mathrm{ext}}})^{N+1}_{\alpha\beta}(c^{\mathrm{ext}})^{N+1}_{N+1,N+1}=(c^{\mathrm{ext}})^{N+1}_{\alpha,N+1}(c^{\mathrm{ext}})^{N+1}_{\beta,N+1}-\sum_{\mu=1}^N c^\mu_{\alpha\beta}(c^{\mathrm{ext}})^{N+1}_{\mu,N+1}.$$ Since the function $F^o_{A_N}$ satisfies  and, as we have just proved, $(c^{\mathrm{ext}})^{N+1}_{\gamma,N+1}=\frac{\d^2 F^o_{A_N}}{\d t^\gamma\d t^{N+1}}$ for $1\le\gamma\le N+1$, we obtain $(c^{\mathrm{ext}})^{N+1}_{\alpha\beta}=\frac{\d^2 F^o_{A_N}}{\d t^\alpha\d t^\beta}$.

$D_N$-case
----------

Consider the space $M^{\mathrm{ext}}_{D_N}:={\mathbb C}^N\times{\mathbb C}^*$ with coordinates $v_1,\ldots,v_{N+1}$. Consider the quotient ring $$\begin{gathered}
{{\widehat{\mathcal{A}}}}^{{\mathrm{ext}}}_{D_N} := {{\mathbb{C}}}[x,y,w,v_1,\dots,v_N,v_{N+1},v_{N+1}^{-1}]\Big/ \left(w - v_{N+1} {{\partial}}_x\Lambda_{D_N},{{\partial}}_y \Lambda_{D_N},2wx - v_{N+1}^2w\right).\end{gathered}$$ As a ${{\mathbb{C}}}[v_1,\dots,v_N,v_{N+1},v_{N+1}^{-1}]$-module, the space ${{\widehat{\mathcal{A}}}}^{{\mathrm{ext}}}_{D_N}$ is free of dimension $N+1$ with a basis $[1],[x],\ldots,[x^{N-2}],[y],[w]$. To show that any other element $[x^i y^j w^k]$ can be expressed in terms of them, first note that $$\begin{gathered}
\label{eq:xy for DN}
[xy]=-\frac{v_N}{2}[1],\qquad[w x]=\frac{v_{N+1}^{2}}{2}[w],\qquad [wy]=-\frac{v_N}{v_{N+1}^2}[w],\end{gathered}$$ where the last equation follows from the first two. Similarly to the $A_N$-case, we have $$\begin{gathered}
\label{eq:w2 for DN}
[w^2]=\left(\frac{v_{N+1}^{2N-3}}{2^{N-2}}+\frac{v_N^2}{v_{N+1}^3}+\sum_{k=2}^{N-1}(k-1)v_k \frac{v_{N+1}^{2k-3}}{2^{k-2}}\right)[w].\end{gathered}$$ Using that $[w-v_{N+1} {{\partial}}_x\Lambda_{D_N}]=0$, we obtain $$\begin{gathered}
\label{eq:y2 for DN}
[y^2]=\frac{1}{v_{N+1}}[w]-[x^{N-2}]-\sum_{k=2}^{N-1}(k-1)v_k[x^{k-2}].\end{gathered}$$ Multiplying this equation by $[x]$, we get the relation $$\begin{gathered}
