Proving a thermodynamic relation between $(\partial H/\partial T)_p$ and $(\partial U/\partial T)_V$
$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$

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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