Proving a thermodynamic relation between $(\partial H/\partial T)_p$ and $(\partial U/\partial T)_V$ (This question is taken from Problem 1.1(b) of the book Chemical Thermodynamics: Principles and Applications.)$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$


  Prove $$\pd HTp = \pd UTV + \left[ V - \pd HpT \right] \pd pTV. \tag{1}$$


First, I expressed $H$ as  $U + PV$, and took the partial derivative against $T$, keeping pressure constant:

$$\pd HTp = \pd UTp + p\pd VTp \tag{2}$$

Expressing $U$ as a function of $(T, V)$:

$$\pd UTp = \pd UTV + \pd UVT \pd VTp \tag{3}$$

Using the well-known identity that $$\pd pVT \pd TpV \pd VTp = -1, \tag{4}$$

$$\pd UVT \pd VTp = -\pd UpT \pd pTV = \left[ V - \pd UpT - V\pd ppT \right]\pd pTV \tag{5}$$

This, of course, means that

$$\pd HTp = \pd UTV + \left[ V - \pd HpT \right]\pd pTV + p\pd VTp \tag{6}$$

Based on my working, there is an additional $$p\pd VTp$$ term. Is my working incorrect? I can’t find my mistake.$\endgroup$
$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$

I would start from

$$ dH = \pd HTp dT + \pd HpT dp$$

This gives rise to 

$$ \pd HTV = \pd HTp  + \pd HpT \pd pTV$$

in which you can recognize some components of the solution. Since

$$dH = dU +PdV + VdP $$ 

so that 

$$ \pd HTV = \pd UTV  + V \pd pTV$$

it follows that 

$$ \pd UTV  + V \pd pTV = \pd HTp  + \pd HpT \pd pTV$$

which is readily rearranged to obtain the desired equation.
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