I am trying to prove that on $\mathbb CP1$ the Fubini-Study form $\omega_{FS}$ is a quarter of the standard form $\omega_{std}= d\theta\wedge dh$ on $S^2$, where we identify $S^2$ with $\mathbb CP^1$ through the stereographic projection.
My take on this:
Let $\psi: S^2\setminus\{N\}\to U_0$ (where $U_0=\{[z_0:z_1]\in\mathbb CP^1: z_o\neq 0\}$) such that $\psi(\theta, h)= [1: \frac{\cos\theta}{1-h}+ i\frac{sin\theta}{1-h}]$. We know that on $[1: x+iy]\in U_0$ the Fubini-Study form is $\omega_{FS}= \frac{dx\wedge dy}{1+x^2+y^2}$. So I tried to pull it back through $\psi$ and prove that $\psi^*\omega_{FS}= \frac14 \omega_{std}$ but I can only show that $\psi^*\omega_{FS}= \frac{d\theta\wedge dh}{(1-h)^2(2-h)}$.
Your formula for the Fubini-Study form is incorrect. The correct formula is $\frac{dx\wedge dy}{(1+x^2+y^2)^2}$. There is a similar formula for the metric, introduced already by Riemann in his 1857 address.