How do I find change in Gibbs free energy of water formation?
You're using the correct equation, $\Delta G^0_r(T;P)=\sum \nu_i \mu_i^0(T;P)$,
where $\Delta G_r^0(T;P)$ is the standard free energy change for the reaction at $T$ and the pressure $P$.  The task is to calculate each $\mu_i^0$.  Provided the same convention for the standard state pressure $P^0$ and for the zero value of $\mu_i$ is used, then $\Delta G_r^0$ for any reaction will be the same.  Here are some ways to calculate $\mu_i^0(T;P)$ for each species to use in the equation and the data required to calculate them (available in tables). In the following, $P^0$ is the standard state pressure used in the tables and subscript $fi$ denotes the reaction forming the species from the elements in their standard states.  If the tabular values are available at a different pressure than the one you want ($P$), you must add $RT\ln(P/P^0)$ to the values of $\mu_i$ indicated below.


$\mu_i^0 = \Delta G^0_{fi}(T;P^0)$.
$\mu_i^0 = \Delta H^0_{fi}(T) - T\Delta S^0_{fi}$.
$\mu_i^0 = \Delta H^0_{fi}(298.15) + \int_{298.15}^Tc_{pi}^0(T) dT -Ts_i^0(T)$.
$\mu_i^0 = \Delta H^0_{fi}(T) - Ts_i^0(T)$.
$\mu_i^0 = \Delta H^0_{fi}(298.15) + \int_{298.15}^Tc_{pi}^0(T)dT -T\left[s_i^0(298.15) + \int_{298.15}^T \frac{c^0_{pi}}{T}dT\right]$


There are many other ways.  It's an interesting exercise to prove why all the above routes give the same value of $\Delta G^0_r$ for any reaction.