In a handful of contexts people study Calabi-Yau threefolds formed by taking the fiber product of two rational elliptic surfaces. I can't find any detailed explanation of why such geometries are actually Calabi-Yau, so I think it's just a straightforward computation which I don't fully understand.

Let $\pi: S \to \mathbb{P}^{1}$ and $\pi' : S' \to \mathbb{P}^{1}$ be two rational elliptic surfaces, and define their fiber product

$$X = S \times_{\mathbb{P}^{1}} S'.$$

With some mild assumptions on $S$ and $S'$, $X$ should be a Calabi-Yau threefold, and I'm hoping someone can help me complete the proof of this. In other words, I want to see that $\omega_{X}=0$ or $K_{X}=0$ (however, note that in general, $X$ will certainly not be smooth).

I believe one should start by considering the obvious map induced by $\pi$ and $\pi'$

$$f: S \times S' \to \mathbb{P}^{1} \times \mathbb{P}^{1}.$$

We can then write $X$ as the pullback of the diagonal $\Delta \subset \mathbb{P}^{1} \times \mathbb{P}^{1}$,

$$X = f^{*} \Delta.$$

So we can realize $X$ as a hypersurface in $S \times S'$, so you should then be able to apply the adjunction formula:

$$\omega_{X} = \omega_{S \times S'}|_{X} \otimes \mathcal{N}_{X/S \times S'},$$

where $\mathcal{N}_{X/S \times S'}$ is the normal bundle of $X$ in $S \times S'$. However, I'm sort of stuck on how to proceed -- How can one explicitly handle both of the two factors in the above tensor product and show they somehow cancel to give 0?