Let $\phi:S^2\to S^2$ be a Möbius transformation. I would like to show that $$\frac1{\sqrt2}|d\phi(z)|=|\phi'(z)|\frac{1+|z|^2}{1+|\phi(z)|^2}.$$ Here I am using the Fubini–Study metric on both domain and codomain. I am pretty unfamiliar with the Fubini–Study metric, but I know it is given by $(dx^2+dy^2)/(1+r^2)^2$.
I tried doing this in local coordinates, but somehow I ended up with the expression $d\phi(z)(v)$ being a real number instead of a tangent vector. I think any explanation would help, but I would really love one in local coordinates so that I can figure out what I was doing wrong. Thanks!