[proofplan] We first shrink the coordinate neighbourhoods so that the hypersurface germs are represented by the actual equality $F(U\cap M)=V\cap N$. This makes $\rho=\sigma\circ F$ vanish exactly on $M$ near $p$, and the non-vanishing of $d\rho_p$ follows from the invertibility of $dF_p$. We then compute the complex Hessian of $\rho$ in standard coordinates; holomorphicity of $F$ removes every term involving an antiholomorphic derivative of $F$. Evaluating this pulled-back mixed Hessian on two complex tangent vectors gives the Levi-form identity. [/proofplan] [step:Shrink the neighbourhoods so that $\rho$ defines $M$ near $p$] Choose open neighbourhoods $U_0\subset U$ of $p$ and $V_0\subset V$ of $q$ such that $F(U_0)=V_0$ and \begin{align*} F(U_0\cap M)=V_0\cap N. \end{align*} This is allowed because the theorem assumes that $F(M)$ and $N$ agree as hypersurface germs near $q$. Since $\sigma:V_0\to\mathbb R$ is a local defining function for $N$, we have \begin{align*} N\cap V_0=\{y\in V_0:\sigma(y)=0\} \end{align*} and $d\sigma_q\ne 0$. Define $\rho:U_0\to\mathbb R$ by $\rho(z)=\sigma(F(z))$. Then, for $z\in U_0$, \begin{align*} \rho(z)=0 \iff F(z)\in N\cap V_0 \iff z\in M\cap U_0. \end{align*} Thus $\rho^{-1}(0)=M\cap U_0$. It remains to check that $d\rho_p\ne 0$. The real differential $dF_p:T_p\mathbb C^n\to T_q\mathbb C^n$ is an isomorphism because $F$ is biholomorphic. By the real chain rule, \begin{align*} d\rho_p=d\sigma_q\circ dF_p. \end{align*} If $d\rho_p=0$, then $d\sigma_q$ would vanish on the range of $dF_p$, which is all of $T_q\mathbb C^n$. This contradicts $d\sigma_q\ne 0$. Hence $d\rho_p\ne 0$, so $\rho$ is a $C^2$ local defining function for $M$ near $p$. [/step] [step:Show that $dF_p$ sends complex tangent vectors on $M$ to complex tangent vectors on $N$] Write the standard coordinates on the source as $z=(z_1,\dots,z_n)$ and on the target as $\zeta=(\zeta_1,\dots,\zeta_n)$. Let $F=(F_1,\dots,F_n)$, where each $F_a:U_0\to\mathbb C$ is holomorphic. For $Z=(Z_1,\dots,Z_n)\in T_{1,0,p}M$, the complex differential of $\rho=\sigma\circ F$ in the direction $Z$ satisfies \begin{align*} \partial\rho_p(Z)=\partial\sigma_q(dF_pZ). \end{align*} Since $Z\in T_{1,0,p}M$ and $\rho$ is a defining function for $M$, $\partial\rho_p(Z)=0$. Therefore \begin{align*} \partial\sigma_q(dF_pZ)=0. \end{align*} By the defining-function characterization of the holomorphic tangent space, this says $dF_pZ\in T_{1,0,q}N$. [guided] We need to verify that the expression $\mathcal L_{\sigma,q}(dF_pZ,dF_pW)$ is legitimate. For that, $dF_pZ$ and $dF_pW$ must lie in $T_{1,0,q}N$. Use the defining-function description of the holomorphic tangent space. Since $\rho$ defines $M$ near $p$, \begin{align*} T_{1,0,p}M=\{Z\in T_{1,0,p}\mathbb C^n:\partial\rho_p(Z)=0\}. \end{align*} Similarly, since $\sigma$ defines $N$ near $q$, \begin{align*} T_{1,0,q}N=\{\Xi\in T_{1,0,q}\mathbb C^n:\partial\sigma_q(\Xi)=0\}. \end{align*} Let $Z=(Z_1,\dots,Z_n)\in T_{1,0,p}M$. Because $F$ is holomorphic, its complex differential maps $(1,0)$ vectors to $(1,0)$ vectors, and in coordinates \begin{align*} (dF_pZ)_a=\sum_{j=1}^n \frac{\partial F_a}{\partial z_j}(p)Z_j. \end{align*} The chain rule for the first complex derivative of $\rho=\sigma\circ F$ gives \begin{align*} \partial\rho_p(Z)=\partial\sigma_q(dF_pZ). \end{align*} The left-hand side is zero because $Z$ is tangent to $M$ in the holomorphic sense. Hence \begin{align*} \partial\sigma_q(dF_pZ)=0. \end{align*} This is exactly the condition that $dF_pZ\in T_{1,0,q}N$. Applying the same argument to $W$ gives $dF_pW\in T_{1,0,q}N$. [/guided] [/step] [step:Compute the mixed complex Hessian of $\rho$ by the holomorphic chain rule] For $1\le j,k,a,b\le n$, denote \begin{align*} \rho_{j\bar k}(p)=\frac{\partial^2\rho}{\partial z_j\partial\bar z_k}(p) \end{align*} and \begin{align*} \sigma_{a\bar b}(q)=\frac{\partial^2\sigma}{\partial \zeta_a\partial\bar\zeta_b}(q). \end{align*} Since $F_a$ is holomorphic, $\partial_{\bar z_k}F_a=0$. Since $\overline{F_b}$ is antiholomorphic, $\partial_{z_j}\overline{F_b}=0$. Applying the complex chain rule to $\rho=\sigma\circ F$ gives \begin{align*} \rho_{j\bar k}(p)=\sum_{a=1}^n\sum_{b=1}^n \sigma_{a\bar b}(q)\frac{\partial F_a}{\partial z_j}(p)\overline{\frac{\partial F_b}{\partial z_k}(p)}. \end{align*} No additional terms occur: the possible second-derivative terms contain either $\partial_{\bar z_k}F_a$ or $\partial_{z_j}\overline{F_b}$, and both are zero by holomorphicity. [/step] [step:Evaluate the pulled-back Hessian on $Z$ and $W$] Using the mixed-Hessian convention for the Levi form, for $Z,W\in T_{1,0,p}M$, \begin{align*} \mathcal L_{\rho,p}(Z,W)=\sum_{j=1}^n\sum_{k=1}^n \rho_{j\bar k}(p)Z_j\overline{W_k}. \end{align*} Substituting the formula for $\rho_{j\bar k}(p)$ gives \begin{align*} \mathcal L_{\rho,p}(Z,W)=\sum_{a=1}^n\sum_{b=1}^n \sigma_{a\bar b}(q)\left(\sum_{j=1}^n\frac{\partial F_a}{\partial z_j}(p)Z_j\right)\overline{\left(\sum_{k=1}^n\frac{\partial F_b}{\partial z_k}(p)W_k\right)}. \end{align*} By the coordinate formula for the complex differential, \begin{align*} (dF_pZ)_a=\sum_{j=1}^n\frac{\partial F_a}{\partial z_j}(p)Z_j \end{align*} and \begin{align*} (dF_pW)_b=\sum_{k=1}^n\frac{\partial F_b}{\partial z_k}(p)W_k. \end{align*} Therefore \begin{align*} \mathcal L_{\rho,p}(Z,W)=\sum_{a=1}^n\sum_{b=1}^n \sigma_{a\bar b}(q)(dF_pZ)_a\overline{(dF_pW)_b}. \end{align*} The right-hand side is precisely \begin{align*} \mathcal L_{\sigma,q}(dF_pZ,dF_pW). \end{align*} Thus \begin{align*} \mathcal L_{\rho,p}(Z,W)=\mathcal L_{\sigma,q}(dF_pZ,dF_pW) \end{align*} for all $Z,W\in T_{1,0,p}M$, completing the proof. [/step]