# Second-order partial derivative of Helmholtz Potential

So the lecture notes to my thermodynamics course contain this relation: $$\left(\frac{\partial^2F}{\partial T^2}\right)_{V,N}=-\frac{1}{\frac{\partial^2U}{\partial S^2}}$$ With no further explanation given. I know that the Helmholtz Potential gives $$\left(\frac{\partial F}{\partial T}\right)_{V,N}=S$$ but how does one arrive at $$\left(\frac{\partial S}{\partial T}\right)_{V,N}=-\frac{1}{\frac{\partial^2U}{\partial S^2}}$$ ?

From the differential expression for $$dF$$, we have: $$\left( \frac{\partial{F}}{\partial{T}}\right)_{V,N} = -S$$. Taking another derivative w.r.t. $$T$$: $$\left( \frac{\partial{F}^2}{\partial{T}^2}\right)_{V,N} = -\left( \frac{\partial{S}}{\partial{T}}\right)_{V,N}.$$
On the other side, from the differential expression for $$dU$$, we have: $$\left( \frac{\partial{U}}{\partial{S}}\right)_{V,N} = T.$$ Taking another derivative w.r.t. $$S$$: $$\left( \frac{\partial{U}^2}{\partial{S}^2}\right)_{V,N} = \left( \frac{\partial{T}}{\partial{S}}\right)_{V,N}.$$ We get the required result recalling that $$\left( \frac{\partial{S}}{\partial{T}}\right)_{V,N} = \frac{1}{ \left( \frac{\partial{T}}{\partial{S}}\right)_{V,N} }$$.
$$\left(\frac{\partial F}{\partial T}\right)_{V,N}=S$$, differentiate this w.r.t Temp keeping constant V,N and then substitute from the first equation.