Since $g_{\mathbb{C}^{3}}$ is a Ricci-flat metric, the Bochner formula implies

$$\Delta(\frac{1}{2}|\nabla u|^{2})=|\nabla^{2}u|^{2}+\langle\Delta\nabla u,\nabla u\rangle=|\nabla^{2}u|^{2}+\langle\nabla\Delta u,\nabla u\rangle=|\nabla^{2}u|^{2},$$

so $|du|^{2}$ is a non-negative subharmonic function. The decay condition implies it converges to zero at infinity, so maximum principle gives $du=0$. ∎

The strategy is to improve the decay rate of $d\phi^{\prime}$ iteratively, until it becomes sufficiently fast. We rewrite the equation as a Poisson equation

$$\frac{1}{2}(\Delta_{g_{\mathbb{C}^{3}}}\phi^{\prime})\omega_{\mathbb{C}^{3}}^{3}=-3(\sqrt{-1}\partial\bar{\partial}\phi^{\prime})^{2}\wedge\omega_{\mathbb{C}^{3}}-(\sqrt{-1}\partial\bar{\partial}\phi^{\prime})^{3}.$$

Notice that $\sqrt{-1}\partial\bar{\partial}\phi^{\prime}$ lives in $C^{k,\alpha}_{\delta,\tau}$, so its square lives in $C^{k,\alpha}_{2\delta,2\tau}$. As long as $(\delta,\tau)$ stays in the good range of weight exponents, Corollary 2.24 and the above vanishing lemma imply that the solution $\phi^{\prime}$ to this Poisson equation must satisfy $\left\lVert d\phi^{\prime}\right\rVert_{C^{k+1,\alpha}_{2\delta+1,2\tau}(\mathbb{C}^{3},\Lambda^{1})}<\infty$. This is an improved decay estimate because $2\delta+1<\delta$ and $2\tau<\tau$. Since each iteration improves the decay rate by a definite amount, within a finite number of steps we can assume $\delta<-2$ and $\tau<0$. Then $\left\lVert d\phi^{\prime}\right\rVert_{C^{k+1,\alpha}_{\delta+1,\tau}(\mathbb{C}^{3},\Lambda^{1})}<\infty$ implies that $\left\lVert\phi^{\prime}\right\rVert_{C^{k+2,\alpha}_{\delta+2,\tau}(\mathbb{C}^{3})}<\infty$ after adjusting $\phi^{\prime}$ by a constant. Then we can use the standard integration by part argument for the complex Monge-Ampère equation to see

$$\int_{\mathbb{C}^{3}}|\nabla\phi^{\prime}|^{2}\phi^{\prime p}\omega_{\mathbb{C}^{3}}^{3}=0,\quad p\gg 1.$$

Hence $\phi^{\prime}$ is a constant, and the metric is unique. ∎