[proofplan] We compute both Levi forms in the same holomorphic coordinate system near $p$. The product rule for the complex Hessian of $\widetilde\rho=a\rho$ produces four terms. At $p\in M$, the term containing $\rho(p)$ vanishes, and the two first-order cross terms vanish because $Z$ and $W$ are complex tangent to $M$. The only remaining contribution is $a(p)$ times the complex Hessian of $\rho$. [/proofplan]

[step:Write the Levi form in local complex coordinates] Use the standard holomorphic coordinates $z=(z_1,\dots,z_n)$ on $U\subset\mathbb C^n$. Write \begin{align*} Z=\sum_{j=1}^n Z_j\partial_{z_j}\big|_p \end{align*} and \begin{align*} W=\sum_{k=1}^n W_k\partial_{z_k}\big|_p \end{align*} for uniquely determined coefficients $Z_j,W_k\in\mathbb C$. For a defining function $r\in C^2(U;\mathbb R)$, the Levi form convention used here is \begin{align*} \mathcal L_{r,p}(Z,W)=\sum_{j=1}^n\sum_{k=1}^n \frac{\partial^2 r}{\partial z_j\partial\bar z_k}(p)Z_j\overline{W_k}. \end{align*} Thus it is enough to compare the complex Hessian coefficients of $\widetilde\rho$ and $\rho$ after pairing with $Z$ and $\overline W$. [/step]

[step:Expand the complex Hessian of the product $a\rho$]For each pair of indices $j,k\in\{1,\dots,n\}$, the ordinary product rule applied first in the $\bar z_k$ direction and then in the $z_j$ direction gives \begin{align*} \frac{\partial^2\widetilde\rho}{\partial z_j\partial\bar z_k}=\frac{\partial^2(a\rho)}{\partial z_j\partial\bar z_k}=a\frac{\partial^2\rho}{\partial z_j\partial\bar z_k}+\frac{\partial a}{\partial z_j}\frac{\partial\rho}{\partial\bar z_k}+\frac{\partial\rho}{\partial z_j}\frac{\partial a}{\partial\bar z_k}+\rho\frac{\partial^2a}{\partial z_j\partial\bar z_k}. \end{align*} Evaluating at $p$ and using $p\in M=\{\rho=0\}$ gives $\rho(p)=0$, hence \begin{align*} \frac{\partial^2\widetilde\rho}{\partial z_j\partial\bar z_k}(p)=a(p)\frac{\partial^2\rho}{\partial z_j\partial\bar z_k}(p)+\frac{\partial a}{\partial z_j}(p)\frac{\partial\rho}{\partial\bar z_k}(p)+\frac{\partial\rho}{\partial z_j}(p)\frac{\partial a}{\partial\bar z_k}(p). \end{align*}[/step]

[guided]We now identify exactly which second-order terms appear when differentiating $\widetilde\rho=a\rho$. Since $a,\rho\in C^2(U;\mathbb R)$, all first and mixed second Wirtinger derivatives appearing below exist and are continuous on $U$. Fix indices $j,k\in\{1,\dots,n\}$. First differentiating in the $\bar z_k$ direction gives \begin{align*} \frac{\partial(a\rho)}{\partial\bar z_k}=a\frac{\partial\rho}{\partial\bar z_k}+\rho\frac{\partial a}{\partial\bar z_k}. \end{align*} Differentiating this identity in the $z_j$ direction and applying the product rule to each product gives \begin{align*} \frac{\partial^2(a\rho)}{\partial z_j\partial\bar z_k}=a\frac{\partial^2\rho}{\partial z_j\partial\bar z_k}+\frac{\partial a}{\partial z_j}\frac{\partial\rho}{\partial\bar z_k}+\frac{\partial\rho}{\partial z_j}\frac{\partial a}{\partial\bar z_k}+\rho\frac{\partial^2a}{\partial z_j\partial\bar z_k}. \end{align*} Because $\widetilde\rho=a\rho$ on $U$, this is the mixed complex Hessian coefficient of $\widetilde\rho$. The point $p$ lies on $M$, and $\rho$ is a defining function for $M$, so $\rho(p)=0$. Therefore the last term disappears after evaluation at $p$, leaving \begin{align*} \frac{\partial^2\widetilde\rho}{\partial z_j\partial\bar z_k}(p)=a(p)\frac{\partial^2\rho}{\partial z_j\partial\bar z_k}(p)+\frac{\partial a}{\partial z_j}(p)\frac{\partial\rho}{\partial\bar z_k}(p)+\frac{\partial\rho}{\partial z_j}(p)\frac{\partial a}{\partial\bar z_k}(p). \end{align*} This is the only place where second derivatives of $a$ could have entered; the defining-function condition $\rho(p)=0$ removes them.[/guided]

[step:Use complex tangency to eliminate the cross terms] Since $Z,W\in T^{1,0}_pM$, the defining function $\rho$ satisfies \begin{align*} \partial\rho_p(Z)=\sum_{j=1}^n \frac{\partial\rho}{\partial z_j}(p)Z_j=0. \end{align*} Because $\rho$ is real-valued, conjugating the tangency identity $\partial\rho_p(W)=0$ gives \begin{align*} \bar\partial\rho_p(\overline W)=\sum_{k=1}^n \frac{\partial\rho}{\partial\bar z_k}(p)\overline{W_k}=0. \end{align*} Pairing the identity from the preceding step with $Z_j\overline{W_k}$ and summing over $j,k$ gives \begin{align*} \mathcal L_{\widetilde\rho,p}(Z,W)=a(p)\mathcal L_{\rho,p}(Z,W)+\left(\sum_{j=1}^n \frac{\partial a}{\partial z_j}(p)Z_j\right)\left(\sum_{k=1}^n \frac{\partial\rho}{\partial\bar z_k}(p)\overline{W_k}\right)+\left(\sum_{j=1}^n \frac{\partial\rho}{\partial z_j}(p)Z_j\right)\left(\sum_{k=1}^n \frac{\partial a}{\partial\bar z_k}(p)\overline{W_k}\right). \end{align*} The second displayed product is zero because $\bar\partial\rho_p(\overline W)=0$, and the third displayed product is zero because $\partial\rho_p(Z)=0$. Therefore \begin{align*} \mathcal L_{\widetilde\rho,p}(Z,W)=a(p)\mathcal L_{\rho,p}(Z,W). \end{align*} [/step]

[step:Conclude the transformation law for all complex tangent vectors] The argument used only that $Z,W\in T^{1,0}_pM$ and that $\widetilde\rho=a\rho$ near $p$. Hence the displayed identity holds for every pair $Z,W\in T^{1,0}_pM$, which is the claimed change-of-defining-function formula for the Levi form. [/step]