Derivative of internal energy with respect to pressure at a constant volume

I need to obtain the derivative of internal energy $$U$$ w.r.t. pressure $$p$$ at a constant volume $$V$$. Realizing that $$\mathrm{d} U = T \mathrm{d} S - p \mathrm{d} V$$, I rewrite $$\left( \frac{\partial U}{\partial p} \right)_V = \left( \frac{\partial U}{\partial S} \right)_V \left( \frac{\partial S}{\partial p} \right)_V$$

The first term is equal to $$T$$ $$\left( \frac{\partial U}{\partial S} \right)_V = T$$

For the second term I used $$\mathrm{d} U = T \mathrm{d} S - p \mathrm{d} V$$ again, from which $$\left( \frac{\partial U}{\partial S} \right)_V = T, \quad \left( \frac{\partial U}{\partial V} \right)_S = - p$$ and therefore $$\left( \frac{\partial p}{\partial S} \right)_V = -\left( \frac{\partial T}{\partial V} \right)_S \quad \implies \quad \left( \frac{\partial S}{\partial p} \right)_V = \frac{1}{\left( \frac{\partial p}{\partial S} \right)_V} = - \frac{1}{\left( \frac{\partial T}{\partial V} \right)_S} = - \left( \frac{\partial V}{\partial T} \right)_S$$

So far, we have $$\left( \frac{\partial U}{\partial p} \right)_V = \left( \frac{\partial U}{\partial S} \right)_V \left( \frac{\partial S}{\partial p} \right)_V = - T \left( \frac{\partial V}{\partial T} \right)_S$$

For $$(\partial V/\partial T)_S$$ I use $$\left( \frac{\partial x}{\partial y} \right)_z \left( \frac{\partial y}{\partial z} \right)_x \left( \frac{\partial z}{\partial x} \right)_y = -1$$ with $$x = V$$, $$y = T$$ and $$z = S$$ $$\left( \frac{\partial V}{\partial T} \right)_S \left( \frac{\partial T}{\partial S} \right)_V \left( \frac{\partial S}{\partial V} \right)_T = -1$$

Using the definition for the heat capacity at constant volume, we get $$C_V = \left( \frac{\partial Q}{\partial T} \right)_V = T \left( \frac{\partial S}{\partial T} \right)_V \quad \implies \quad \left( \frac{\partial T}{\partial S} \right)_V = \frac{1}{\left( \frac{\partial S}{\partial T} \right)_V} = \frac{1}{C_V / T} = \frac{T}{C_V}$$

For $$\left( \frac{\partial S}{\partial V} \right)_T$$ I use Helmholtz free energy $$\mathrm{d} F = - p \mathrm{d} V - S \mathrm{d} T$$ so $$\left( \frac{\partial F}{\partial V} \right)_V = - p, \quad \left( \frac{\partial F}{\partial T} \right)_S = - S$$ therefore $$\left( \frac{\partial S}{\partial V} \right)_T = \left( \frac{\partial p}{\partial T} \right)_V$$ and for the last time $$\left( \frac{\partial p}{\partial T} \right)_V \left( \frac{\partial T}{\partial V} \right)_p \left( \frac{\partial V}{\partial p} \right)_T = -1$$ with $$\alpha = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_p \quad \implies \quad \left( \frac{\partial T}{\partial V} \right)_p = \frac{1}{\left( \frac{\partial V}{\partial T} \right)_p} = \frac{1}{\alpha V}$$ and $$K_T = - \frac{1}{V} \left( \frac{\partial V}{\partial p} \right)_T \quad \implies \quad \left( \frac{\partial V}{\partial p} \right)_T = - K_T V$$ and therefore $$\left( \frac{\partial S}{\partial V} \right)_T = \left( \frac{\partial p}{\partial T} \right)_V = - \frac{1}{\left( \frac{\partial T}{\partial V} \right)_p \left( \frac{\partial V}{\partial p} \right)_T} = \frac{\alpha}{K_T}$$ going back to $$\left( \frac{\partial V}{\partial T} \right)_S = - \frac{1}{\left( \frac{\partial T}{\partial S} \right)_V \left( \frac{\partial S}{\partial V} \right)_T} = - \frac{1}{(T/C_V) (\alpha/K_T)} = - \frac{C_V K_T}{\alpha T}$$ and going back to the internal energy... $$\left( \frac{\partial U}{\partial p} \right)_V = - T \left( \frac{\partial V}{\partial T} \right)_S = - T \left( - \frac{C_V K_T}{\alpha T} \right) = \frac{C_V K_T}{\alpha}$$

However, the formula should be $$\left( \frac{\partial U}{\partial p} \right)_V = \frac{C_p K_T}{\alpha} - \alpha V T$$

My question is: how do I obtain the second formula as quickly as possible without using Mayer's relation $$C_p - C_V = (\alpha^2/K_T) V T$$?

P.S. I also found a quicker way to obtain the formula with $$C_V$$ $$\left( \frac{\partial U}{\partial p} \right)_V = \left( \frac{\partial U}{\partial T} \right)_V \left( \frac{\partial T}{\partial p} \right)_V = C_V \frac{K_T}{\alpha} = \frac{C_V K_T}{\alpha}$$

$$dH=dU+pdV+Vdp=C_pdT+V(1-\alpha T)dP$$So, $$dU=C_pdT-pdV-V\alpha Tdp$$The rest is basically what you've already done with your second method.