Let $Y$ be a closed Kähler manifold with $c_1(Y)=0$ in $H^2(Y,\mathbb{R})$. Let $\omega$ be a Ricci-flat Kähler form on $\mathbb{C}^m \times Y$ such that $$A^{-1} (\omega_{\mathbb{C}^m} + \omega_Y) \leq \omega \leq A  (\omega_{\mathbb{C}^m} + \omega_Y),$$ for some constant $A \geq 1$, where $\omega_Y$ is a Kähler form on $Y$ and $\omega_{\mathbb{C}^m}$ is the Euclidean form on $\mathbb{C}^m$. 

I want to show that there is a unique choice of $\omega_Y$ such that $\text{Ric}(\omega_Y)=0$ and that there is a smooth function $f$ such that $$\omega = \omega_{\mathbb{C}^m} + \omega_Y + d f.$$


_Update:_ I can show that the Kähler class of $\omega$ is independent of the $\mathbb{C}^m$ component by using Künneth's formula. To establish the existence of $f$ I have been attempting to write out the components of $\omega$ as a section of $\Lambda^2(\mathbb{C}^m \times Y) = \Lambda^2(\mathbb{C}^m) \oplus \Lambda^1(\mathbb{C}^m) \otimes \Lambda^1(Y) \oplus \Lambda^2(Y)$ and using the fact that $d\omega=0$.

Indeed, let $\vartheta_b = \vartheta_{b^1}, ..., \vartheta_{b^m}$ be a basis of $(1,0)$-forms on $\mathbb{C}^m$, such that $\overline{\vartheta}_b = \overline{\vartheta}_{b^1}, ..., \overline{\vartheta}_{b^m}$ is a basis of $(0,1)$-forms on $\mathbb{C}^m$. Similarly, let $\vartheta_h = \vartheta_{h^1}, ..., \vartheta_{h^n}$ be a basis of $(1,0)$-forms on $Y$, such that $\overline{\vartheta}_h = \overline{\vartheta}_{h^1}, ..., \overline{\vartheta}_{h^n}$ is a basis of $(0,1)$-forms on $Y$. Then write $$\omega = \frac{\sqrt{-1}}{2}\left( \alpha \vartheta_v \wedge \overline{\vartheta}_v + \beta \vartheta_h \wedge \overline{\vartheta}_v + \overline{\beta} \vartheta_v \wedge \overline{\vartheta}_h + \gamma \vartheta_h \wedge \overline{\vartheta}_h \right).$$

Then \begin{eqnarray*}
d\omega &=& (d_h \alpha) \vartheta_h \wedge \vartheta_v \wedge \overline{\vartheta}_v + (d_h\alpha) \overline{\vartheta}_h \wedge \vartheta_v \wedge \overline{\vartheta}_v \\
&& + (d_h \beta) \overline{\vartheta}_h \wedge \vartheta_h \wedge \overline{\vartheta}_v + (d_v \beta) \vartheta_v \wedge \vartheta_h \wedge \overline{\vartheta}_v \\
&& + (d_v \overline{\beta}) \overline{\vartheta}_v \wedge \vartheta_v \wedge \overline{\vartheta}_h + (d_h \overline{\beta}) \vartheta_h \wedge \vartheta_v \wedge \overline{\vartheta}_h + (d_v \gamma) \vartheta_v \wedge \vartheta_h \wedge \overline{\vartheta}_h \\
&& + (d_v \gamma) \overline{\vartheta}_v \wedge \vartheta_h \wedge \overline{\vartheta}_h
\end{eqnarray*}



_Another attempt:_ Consider the function $$\varphi : = \log \left( \frac{\omega^{m+n}}{\omega_{\mathbb{C}^m} \wedge \omega_Y^n} \right).$$ It is obvious from the fact that $\omega$ and $\omega_{\mathbb{C}^m} + \omega_Y$ are Ricci-flat that $\varphi$ is pluriharmonic, i.e., $\sqrt{-1}\partial \overline{\partial} \varphi =0$. Observe that \begin{eqnarray*}
0  = \sqrt{-1} \partial \overline{\partial} \varphi &=& \sqrt{-1} \partial \overline{\partial} \log(\omega^{m+n}) - \sqrt{-1} \partial \overline{\partial} \log(\omega_{\mathbb{C}^m} \wedge \omega_Y^n) \\
&=& - \text{Ric}(\omega) - \text{Ric}(\omega_{\mathbb{C}^m} + \omega_Y).
\end{eqnarray*}
Hence, we see that $\text{Ric}(\omega) = \text{Ric}(\omega_{\mathbb{C}^m} + \omega_Y)$. 

Now (if I recall), if $\omega$ is a Kähler form, the Ricci form $\text{Ric}(\omega)$ lies in the same de Rham cohomology class as $\omega$. Thus, $\text{Ric}(\omega) - \omega = df$ for some smooth real valued function $f$. Similarly, there is a smooth real valued function $\tilde{f}$ such that $\text{Ric}(\omega_{\mathbb{C}^m} + \omega_Y) - (\omega_{\mathbb{C}^m} + \omega_Y) = d\tilde{f}$. Hence, \begin{eqnarray*}
\text{Ric}(\omega) = \text{Ric}(\omega_{\mathbb{C}^m} + \omega_Y)  & \implies & \omega + df = \omega_{\mathbb{C}^m} + \omega_Y + d \tilde{f},
\end{eqnarray*}

and by setting $f : = \tilde{f} - f$, we have the desired result. 

_Potential Issue:_ Of course, in the Kähler-Einstein case, the Kähler form and the Ricci form always lie in the same de Rham cohomology class. Moreover, since we are talking about Calabi-Yau manifolds, we are in the Kähler-Einstein case. I still need to check this argument in case I have missed something, however. (If someone can point out any technicalities I might have overlooked, that would be greatly appreciated. I would also appreciate a more direct approach like the one I originally tried.