Differentiate Gibbs free energy maybe by using chain rule? I found in a book this function $ G(P,T,n_1,...) $ and its differential: $$ dG = \left( \frac{\partial G}{\partial P} \right)_{T, n_1,...} dt + \left( \frac{\partial G}{\partial T} \right)_{P, n_1,...} dt +...$$ The pedices mean that those variables are considered constants. Can you tell me how can I get the above result please? I know only that if I have the function $ f(x,y) $, the differential is: $$ df = \left( \frac{\partial f}{\partial x} \right) dx + \left( \frac{\partial f}{\partial y} \right) dy $$ but I think that this is another story (Taylor stuff). Thank you in advance.

Don't worry, that is the same story! The Gibbs free energy is a thermodynamic potential which is a function of $T,p,N_{1},...,N_{k}$, where $T$ is the temperature of the system, $p$ is its pressure and $N_{1},...,N_{k}$ are the number of particles of each constituent of the system. For simplicity, let us consider $k=1$ and $N_{1} \equiv N$. Then $G = G(T,p,N)$. Now, by your definition of a differential, we have:

$$ dG = \bigg{(}\frac{\partial G}{\partial T}\bigg{)}dT + \bigg{(}\frac{\partial G}{\partial p}\bigg{)}dp + \bigg{(}\frac{\partial G}{\partial N}\bigg{)}dN $$