I would like to know if the function $$\frac{1}{(1+\vert z\vert^2)^2}$$ is subharmonic on $\mathbf{C}=\mathbf{P}^1-\{\infty\}$.
Motivation. The Fubini-Study metric on the complex projective line $\mathbf{P}^1$ is given by $$\frac{i}{2\pi} \frac{dz \wedge d\overline{z}}{(1+\vert z\vert^2)^2} ;$$ this depends on the choice of the coordinate $z$.
I can think of two possiblities. Firstly, just an ugly computation with derivatives. Secondly, an argument using properties of the Fubini-study metric, i.e., it is a smooth positive real $(1,1)$-form.
Question. Is the above function subharmonic?
Question. If yes, then is it possible to prove this without calculating derivatives?
No, this function is not subharmonic on the whole plane, as an easy computation can show (its $\partial \bar \partial$ equals something like $2(2|z|^2-1)/(1+|z|^2)^4$.)
As for the Fubini-Study metric, it can be viewed as the curvature form of the so-called Fubini-Study metric on the line bundle $\mathcal O_{\mathbb P^1}(1)$, its local potential on $\mathbb P^1 - \{\infty\}$ is given by $\varphi(z)= \log (1+|z|^2)$, so that the metrics is $i \partial \bar \partial \varphi$, which is well-defined globally. Its local expression is then as you wrote
$$\omega_{\rm FS}=\frac{i}{2\pi} \frac{dz\wedge d\bar z}{(1+|z|^2)^2}$$
To sum up, whenever you have a positive (1,1) form $\omega$ on a manifold, then its local potentials (ie the functions $\varphi$ such that $\omega = i\partial \bar \partial \varphi$ locally) are (pluri)subharmonic functions, but you can't say anything about the term in $dz \wedge d\bar z$, except that it is positive (or in higher dimension, the terms in $dz_i \wedge d\bar z_j$ form a positive definite hermitian matrix).