Deriving heat capacity in terms of internal energy U and natural variables S & V
Usually this part of thermodynamics is not presented in the most efficient way. To derive the expression for, e.g., $C_P - C_V$, one goes through a series of steps involving Maxwell's relations and the triple product rule. Figuring out what to do in each step is done mostly by guesswork.

However, all relations of these sorts really come from just a single mathematical relation that more generally holds. That is to say,


  Consider a twice-differentiable,  two-variable function $f(x,y)$ and its Legendre transform with respect to $y$, defined by $g(x,f_y) := f - f_y y$. 1  Here, $f_y$ denotes the derivative of $f$ with respect of $y$ with other independent variables held fixed. Then, the following identity is satisfied:
  \begin{equation}
\boxed{g_{xx}  - f_{xx}  = - \frac{f_{xy}^2}{f_{yy}} = \frac{g_{x f_y}^2}{g_{f_y f_y}}}.
\end{equation}
  Proof. We have 
  \begin{equation}
\bigg(\frac{\partial f}{\partial x} \bigg)_{f_y} = f_x + f_y \bigg(\frac{\partial y}{\partial x}\bigg)_{f_y} = f_x - f_y \frac{f_{xy}}{f_{yy}}.
\end{equation}
  On the above, the first equality holds because of the chain rule [for the change of variables $(x,y)$ $\to$ $(x, f_y)$], and the second equality because of the triple product rule. Then,
  \begin{equation}
g_x = \bigg(\frac{\partial f}{\partial x} \bigg)_{f_y} - f_y\bigg(\frac{\partial y}{\partial x}\bigg)_{f_y}   =f_x,
\end{equation}
  and
  \begin{equation}
g_{xx} = \bigg(\frac{\partial f_x}{\partial x} \bigg)_{f_y}= f_{xx} + f_{xy}\bigg(\frac{\partial y}{\partial x}\bigg)_{f_y} = f_{xx} - \frac{f_{xy}^2}{f_{yy}}.
\end{equation}
  Also, using the chain rule for the change of variables $(x,y)$ $\to$ $(x, f_y)$, we get
  \begin{equation}
g_{f_y} = \bigg(\frac{\partial f}{\partial f_y} \bigg)_{x} -\bigg(\frac{\partial y}{\partial f_y} \bigg)_{x}f_y - y = -y.
\end{equation}
  Therefore, Legendre transform of $g$ with respect to $f_y$ is simply $g - g_{f_y} f_y = f$. Then, performing a similar analysis as above for $g$ gives
  \begin{equation}
f_{xx} = g_{xx} - \frac{g_{x f_y}^2}{g_{f_y f_y}}.
\end{equation}


Now, all that remains to be done is to choose the variables and functions $x$, $y$, $f$ and $g$. Below are some examples.

(1) $x = S$, $y = V$, $f = U$, $g = H$, and $f_y = -P$:
\begin{equation}
H_{SS} - U_{SS} = -\frac{U_{SV}^2}{U_{VV}}.
\end{equation}
Note that
\begin{equation}
U_{SS} = \bigg(\frac{\partial^2 U}{\partial S^2}\bigg)_V = \bigg(\frac{\partial T}{\partial S}\bigg)_V = \frac{T}{T(\partial S/\partial T)_V} = \frac{T}{C_V},
\end{equation}
and that, similarly,
\begin{equation}
H_{SS} = \bigg(\frac{\partial^2 H}{\partial S^2}\bigg)_P = \bigg(\frac{\partial T}{\partial S}\bigg)_P = \frac{T}{T(\partial S/\partial T)_P} = \frac{T}{C_P}.
\end{equation}
Hence, we have
\begin{equation}
\boxed{\frac{1}{C_P} - \frac{1}{C_V} = -\frac{U_{SV}^2}{T U_{VV}} = -\frac{\Big(\frac{\partial^2 U}{\partial S \partial V}\Big)^2}{\Big(\frac{\partial U}{\partial S}\Big)_V \Big(\frac{\partial^2 U}{\partial V^2}\Big)_S}}
\end{equation}
which is equivalent to the relation OP wanted to derive.

(2) $x = T$, $y = V$, $f = A$, $g = G$, $f_y = -P$, and noting that $f_{-y} = -f_y$,
\begin{equation}
G_{TT} - A_{TT} = \frac{G_{T,-P}^2}{G_{-P,-P}} = \frac{G_{TP}^2}{G_{PP}}.
\end{equation}
Here,
\begin{equation}
A_{TT} = \bigg(\frac{\partial^2 A}{\partial T^2}\bigg)_V = -\bigg(\frac{\partial S}{\partial T}\bigg)_V = - \frac{1}{T}C_V,
\end{equation}
\begin{equation}
G_{TT} = \bigg(\frac{\partial^2 G}{\partial T^2}\bigg)_P = -\bigg(\frac{\partial S}{\partial T}\bigg)_P = - \frac{1}{T}C_P,
\end{equation}
\begin{equation}
G_{TP} = \frac{\partial^2 G}{\partial T \partial P} = \bigg(\frac{\partial V}{\partial T}\bigg)_P = V \alpha,
\end{equation}
where $\alpha := \frac{1}{V}\Big(\frac{\partial V}{\partial T}\Big)_P$ is the volume expansion coefficient,
\begin{equation}
G_{PP} = \bigg(\frac{\partial^2 G}{\partial P^2}\bigg)_T = \bigg(\frac{\partial V}{\partial P}\bigg)_T = - V\kappa_T,
\end{equation}
where $\kappa_T := -\frac{1}{V}\Big(\frac{\partial V}{\partial P}\Big)_T$ is the isothermal compressibility. Then, it follows that
\begin{equation}
\boxed{C_p - C_V = \frac{TV\alpha^2}{\kappa_T}}.
\end{equation}

(3) $x = V$, $y = S$, $f = H$, $g = G$, and $f_y = T$:
\begin{equation}
G_{PP} - H_{PP} = \frac{G_{TP}^2}{G_{TT}}.
\end{equation}
We have
\begin{equation}
H_{PP} = \bigg(\frac{\partial^2 H}{\partial P^2}\bigg)_S = \bigg(\frac{\partial V}{\partial P}\bigg)_S = - V\kappa_S,
\end{equation}
where $\kappa_S := -\frac{1}{V}\Big(\frac{\partial V}{\partial P}\Big)_S$ is the adiabatic compressibility. We have already shown that
\begin{equation}
G_{PP} = -V\kappa_T, \quad G_{TP} = V \alpha, \quad G_{TT} = - \frac{C_P}{T},
\end{equation}
whence we obtain
\begin{equation}
\boxed{\kappa_T - \kappa_S = \frac{TV\alpha^2}{C_P}}.
\end{equation}



1 To be more precise, one should first write down $f(x,y) - y f_y$ and then express $y$ as a function of $x$ and $f_y$. Also, for the Legendre transform to exist, $f(x,y)$ should be a convex or concave function of $y$.