On $\mathbb{CP}^n$, we have $\phi_{\alpha}([z^1,...z^{n+1}])=(\omega_{\alpha}^1,...,\omega_{\alpha}^n)$ where $$\omega_{\alpha}^i=\begin{cases} \frac{z^i}{z^{\alpha}}, & \text{if $1\leqslant i \lt \alpha$ } \\ \frac{z^{i+1}}{z^{\alpha}}, & \text{if $\alpha \leqslant i \leqslant n$} \end{cases}$$
The Fubini-Study metric is defined as $$h|_{U_{\alpha}}=\frac{\partial^2 ln f_{\alpha}}{\partial \omega_{\alpha}^i \partial \overline {\omega_{\alpha}^j}} d\omega_{\alpha}^i \otimes d\overline {\omega_{\alpha}^j}$$ where $f_{\alpha}=1+\sum_{i=0}^n |\omega_{\alpha}^i|^2$
I want to show that on $U_{\alpha} \cap U_{\beta}$, we still have $h|_{U_{\alpha}}=h|_{U_{\beta}}$. More precisely, we have $$\frac{\partial^2 ln f_{\alpha}}{\partial \omega_{\alpha}^i \partial \overline {\omega_{\alpha}^j}} d\omega_{\alpha}^i \otimes d\overline {\omega_{\alpha}^j}=\frac{\partial^2 ln f_{\beta}}{\partial \omega_{\beta}^i \partial \overline {\omega_{\beta}^j}} d\omega_{\beta}^i \otimes d\overline {\omega_{\beta}^j}$$
I have already showed that on $U_{\alpha} \cap U_{\beta}$, we have $f_{\alpha}=f_{\beta} |\omega_{\alpha}^{\beta}|^2$. So we have $$\frac{\partial^2 ln f_{\alpha}}{\partial \omega_{\alpha}^i \partial \overline {\omega_{\alpha}^j}}=\frac{\partial^2 ln f_{\beta}}{\partial \omega_{\alpha}^i \partial \overline {\omega_{\alpha}^j}}+\frac{\partial^2 ln \omega_{\alpha}^{\beta}}{\partial \omega_{\alpha}^i \partial \overline {\omega_{\alpha}^j}}+\frac{\partial^2 ln \overline{\omega_{\alpha}^{\beta}}}{\partial \omega_{\alpha}^i \partial \overline {\omega_{\alpha}^j}}$$
That's what I have tried but I am unable to continue. Can anyone else give me some help? Appreciate your feedback.
First, you MISS the coefficent $\frac{i}{2\pi}$ when defining the Fubini-Study metric. And $f_{\alpha} \,\text{should be}\,1+\sum_{i=1}^n |\omega_{\alpha}^i|^2$ rather than $1+\sum_{i=0}^n |\omega_{\alpha}^i|^2.$
You want to show $f_\alpha|_{U\alpha\cap U\beta}=f_\beta|_{U\alpha\cap U\beta}$, i.e. $$\frac{i}{2\pi}\partial\bar\partial\log \left(\sum_{l=1}^{n+1}{|\frac{z_l}{z_\alpha}|}^2\right)=\frac{i}{2\pi}\partial\bar\partial\log \left(\sum_{l=1}^{n+1}{|\frac{z_l}{z_\beta}|}^2\right).$$ Indeed, $$\log \left(\sum_{l=1}^{n+1}\left|\frac{z_{l}}{z_{\alpha}}\right|^{2}\right)=\log \left(\left|\frac{z_{\beta}}{z_{\alpha}}\right|^{2} \sum_{l=1}^{n+1}\left|\frac{z_{l}}{z_{\beta}}\right|^{2}\right)=\log \left(\left|\frac{z_{\beta}}{z_{\alpha}}\right|^{2}\right)+\log \left(\sum_{l=1}^{n+1}\left|\frac{z_{l}}{z_{\beta}}\right|^{2}\right).$$
Thus, we just need to verify $\partial\bar\partial\log\left(|\frac{z_\beta}{z_\alpha}|^2\right)=0$ on $U_\alpha\cap U_\beta$, that is obvious thanks to the fact that for every $f\in\mathcal O^*(U)$ , $\log |f|$ is pluriharmonic on $U$, i.e. $$\partial\bar\partial \log|f|=0,$$where $U$ is a open subset in a complex manifold.