Work with thermodynamical equilibrium condition A thermodynamic system being in thermodynamic equilibrium
is characterized by the property that for every
thermodynamic potential $F$ which describes the system, its differential $dF$ is zero. Let consider for example the internal energy $U(S, V, N_i)$ for now. If the system is in thermodynamic equilibrium then
$$dU= \frac{\partial U}{\partial S}dS + 
\frac{\partial U}{\partial V}dV +
\sum_i^n \frac{\partial U}{\partial N_i}dN_i = TdS + 
pdV +
\sum_i^n \mu_i dN_i = 0 $$
Note that $\frac{\partial U}{\partial S} =T, 
\frac{\partial U}{\partial V} =p, 
\frac{\partial U}{\partial N_i} =\mu_i$.
Question: How this condition $dU$ helps "in practice" when one works with concrete systems and want to find out in which $(S_0, V_0, (N_i)_0)$ the system has its
"equilibrium"?
When I try to apply it I obtain something nonsensical and I want to understand which mistake I make here. Back to our condition $dU=0$ implies that $\frac{\partial U}{\partial S} =T, \frac{\partial U}{\partial V} =p, \frac{\partial U}{\partial N_i} =\mu_i$ should be all zero, because  $S, V $ and $N_i$ are independent variables, therefore the differentials $dS, dV$ and $dN_i$ as well. So I obtain $n+2$ conditions
$\frac{\partial U}{\partial S} =0, 
\frac{\partial U}{\partial V} =0, 
\frac{\partial U}{\partial N_i} =0$
But this not make any sense to me simply because this would imply that if the state is equilibrium state, then always its $T, p $ and $\mu$ are all zeroes. But certainly there are thermodynamical systems which are in equilibrium but their $T, p, \mu_i$ are not zero.
I'm confused now, what I'm doing wrong? could anybody explain to me how to "read"
and "work" with the condition $dU=0$ correctly? sorry, if my question is too easy for people with elementary knowledge on this topic but also after long search I nowhere found an answer.
 A: You have been misled by ab incomplete statement of the equilibrium condition. The precise statement is that for a system at thermodynamic equilibrium under the condition of a fixed set of thermodynamic variables, the corresponding thermodynamic potential is minimum with respect to any additional variable representing a possible internal constraint.
In practice, this means that if you have an isolated system characterized by fixed values of entropy, volume, and number of molecules, its internal energy $U(S,V,N)$ at equilibrium is minimum with respect to any other variable different from those determining the thermodynamic state. For example, if the system is in a container and a fixed, impenetrable and insulating wall is separating subsystem $1$ from subsystem $2$, this is equivalent to have two separate subsystems  with energies $U_1(S_1,V_1,N_1)$ and $U_2(S_2,V_2,N_2)$. If the constraint on thermal insulation is relaxed, and heat can flow between the two subsystems varying $S_1$ and $S_2$ but  without  entropy production, $S_1+S_2=S$, then only one additional independent variable, say $S_1$,  represents the constraint. The vanishing of the first order variation of the total energy with respect to $S_1$
$$
\frac{\partial{U(S,V,N,S_1)}}{\partial{S_1}}=0,
$$
at fixed $S,V,N$, provides the condition for thermal equilibrium after removal of the constraint, i.e. the equality of the temperatures of the two subsystems.
