Deriving an expression for the constant pressure heat capacity
You could start by dividing your whole equation by $T$:

\begin{align}
 \frac{G}{T} = \frac{H}{T} - S
\end{align}

and then take the derivative with respect to $T$ at constant $p$:

\begin{align}
 \left(\frac{\partial\frac{G}{T}}{\partial T}\right)_{p} = \left(\frac{\partial\frac{H}{T}}{\partial T}\right)_{p} - \left(\frac{\partial S}{\partial T} \right)_{p} \ .
\end{align}

That way you've already introduced the desired derivative of the entropy with respect to $T$ at constant $p$.
Then you can use that

\begin{align}
 \frac{G}{T} = \frac{H}{T} - S  \qquad &amp;\Rightarrow&amp; \qquad \mathrm{d} \!\left(\frac{G}{T}\right) &amp;=  - \frac{H}{T^2} \mathrm{d} T + \frac{1}{T} \underbrace{\mathrm{d} H}_{= T \mathrm{d}S + V\mathrm{d}p} - \mathrm{d} S \\
&amp;&amp;&amp;= - \frac{H}{T^2} \mathrm{d} T + \frac{V}{T} \mathrm{d}p \\
&amp;\Rightarrow&amp; \qquad \left(\frac{\partial\frac{G}{T}}{\partial T}\right)_{p} &amp;= - \frac{H}{T^2}
\end{align}

and simply using the product rule

\begin{align}
 \left(\frac{\partial\frac{H}{T}}{\partial T}\right)_{p} = \frac{1}{T} \underbrace{\left(\frac{H}{\partial T}\right)_{p}}_{= C_{p}} + H \underbrace{\left(\frac{\partial \frac{1}{T}}{\partial T} \right)_{p}}_{=-\frac{1}{T^2}} = \frac{C_{p}}{T} - \frac{H}{T^2} \ .
\end{align}

Putting it all together gives you the desired result:

\begin{align}
 \left(\frac{\partial\frac{G}{T}}{\partial T}\right)_{p} &amp;= \left(\frac{\partial\frac{H}{T}}{\partial T}\right)_{p} - \left(\frac{\partial S}{\partial T} \right)_{p} \\
- \frac{H}{T^2} &amp;= \frac{C_{p}}{T} - \frac{H}{T^2} - \left(\frac{\partial S}{\partial T} \right)_{p} \\
\Rightarrow \left(\frac{\partial S}{\partial T} \right)_{p} &amp;= \frac{C_{p}}{T} \ .
\end{align}