Let $\phi(z)$ be a Maass form of type $(v_1,v_2)\in \mathbb{C}^2$ for $SL(3,\mathbb{Z})$. Then the dual Maass form $$ \tilde{\phi}(z):= \phi(w.(z^{-1})^{\intercal}.w)\,,\,\,\,\,\,\,\,\,\,\,w=\begin{pmatrix} & & 1 \\ & -1& \\ 1 & \end{pmatrix}$$ is a Maass form of type $(v_2,v_1)$ .
I am having difficulty showing the part $$\Delta_1\tilde{\phi}=\lambda_1\tilde{\phi}\\\Delta_2\tilde{\phi}=\lambda_2\tilde{\phi} $$ where $\Delta_1, \Delta_2$ are the generators of the centre of the ring of differential operators and $\lambda_1$ and $\lambda_2$ are the eigenvalues of $\Delta_1$ and $\Delta_2$ respectively corresponding to the eigenfunction $I_{v_2,v_1}(z)$ .
Note that if
$$z=\begin{pmatrix}
1 & x_2 & x_3\\
& 1 & x_1\\
& & 1
\end{pmatrix}.\begin{pmatrix}
y_1y_2 & & \\
& y_1 & \\
& & 1
\end{pmatrix}$$ then
$$w.(z^{-1})^{\intercal}.w=\begin{pmatrix}
1 & x_1 & x_1x_2-x_3\\
& 1 & x_2\\
& & 1
\end{pmatrix}.\begin{pmatrix}
y_2y_1 & & \\
& y_2 & \\
& & 1
\end{pmatrix}$$ and hence we have
$$I_{v_1,v_2}(z)={y_1}^{v_1+2v_2}{y_2}^{2v_1+v_2}=I_{v_2,v_1}(w.(z^{-1})^{\intercal}.w)$$ I am not able to see the 'chain rule' the text is then suggesting to use for the conclusion.