Can we show that the helmholtz free energy at equilibrium is minimized from its second derivative? I can show that $dF = 0$ at equilibrium, where F is Helmholtz free energy. But mathematically, starting from $$dF = d(U-TS)$$ I want to show that its second derivative, $d^2F > 0$ at equilibrium and thus I can prove that the equilibrium state has minimum Gibbs free energy. I am starting like $$d^2F=d(d(U-TS))$$ $$d^2F = d(dU-TdS-SdT)$$$$d^2F=d(TdS-PdV-TdS-SdT)$$$$d^2F=d(-PdV-SdT)$$ For a system at constant volume, $dV =0$ and at equilibrium state, $dT =0 $, so clearly I would get $d^2F=0$ which is obviously incorrect, so am I doing something wrong in here, or my approach is totally wrong. Please do suggest. Thanks.
 A: The first derivative at an extremum point $x_0$ is $0$ only when evaluated at $x_0$. Same goes with the second derivative. So you first have to evaluate the second derivative, then you plug in the specific $x=x_0$. 
For example, take $y = x^2$. $\frac{\mathrm{d}y}{\mathrm{d}x}= 2x $ and $\frac{\mathrm{d}^2y}{\mathrm{d^2}x}= 2 $. At $x = 0$, the minimum, $\frac{\mathrm{d}y}{\mathrm{d}x} |_0 = 0$, and $\frac{\mathrm{d}^2y}{\mathrm{d^2}x}|_0= 2$.  If we were to apply your logic however, we would have $\frac{\mathrm{d}^2y}{\mathrm{d^2}x} = \frac{\mathrm{d}}{\mathrm{d}x} \underbrace{\left (\frac{\mathrm{d}y}{\mathrm{d}x}  \right)}_{=0}=0.$
So, you have to expand your last line.
$$\mathrm{d}^2F = -\mathrm{d}P\mathrm{d}V - P\mathrm{d}^2V-\mathrm{d}S\mathrm{d}T-S\mathrm{d}^2T. $$
Now you apply the condition that you are at equilibrium, so $\mathrm{d}T|_{\mathrm{eq}} = \mathrm{d}V|_{\mathrm{eq}} = \mathrm{d}P|_{\mathrm{eq}} = \mathrm{d}S|_{\mathrm{eq}} = 0$, so that:
$$ \mathrm{d}^2F|_{\mathrm{eq}} = - P\mathrm{d}^2V|_{\mathrm{eq}}  -S\mathrm{d}^2T|_{\mathrm{eq}} .  $$
Then I guess that if pressure is positive, then the volume is a maximum so $\mathrm{d}^2V<0$ so that the first term is positive. Probably a similar argument for the second term...?
A: Start with the 1st and 2nd Laws written as $dU=\delta Q + \delta W$, $F=U-TS$ and $\delta Q \le TdS$, where $\delta W$ is the work done by external forces on the system whose internal energy is $U$, and $\delta Q$ is the heat transferred from the environment to the system. 
Then we have for any process that $$d(F+TS)=dF+TdS+SdT \le TdS + \delta W $$ and 
$$dF+SdT \le \delta W \tag{1}\label{1}$$
where in $\eqref{1}$ equality holds iff the process is reversible. 
Now assume that the process is such that the external work done is zero $\delta W=0$ then you have $dF\vert_{\delta W=0} \le -SdT$, and if the process is also isothermal, that is $T=const,\; dT=0$, then you must have $$dF \le 0 \tag{2}\label{2}$$
What does it mean that the free energy cannot increase $\eqref{2}$? If a reversible process then $dF=0$ and it is not changing. If it is an irreversible process then $dF<0$ and $F$ must decrease. As $F$ is bounded from below when decreases it must reach a minimum eventually. Being a minimum its 2nd derivative, if exists, must be positive, that is equilibrium with the given constraints.

Important: Actually, we do not have to assume that the whole process is isothermal. Instead, it is sufficient to assume that the heat (entropy) exchange with the environment is always at the same temperature, $dT=0$. For example, the system's internal temperature during equilibration may change, only its interaction with the outside world must be at a fixed temperature.

