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.)$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)\_{\!#3}}$



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