For this problem it looks like we’ll have three 1^st order partial derivatives to compute. Here they are,

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}\frac{{\partial z}}{{\partial p}} & = {z_p} = \frac{{2p\left( {r + 1} \right)}}{{{t^3}}} + r\,{{\bf{e}}^{2p + 3r + 4t}} + 2pr\,{{\bf{e}}^{2p + 3r + 4t}}\\ \frac{{\partial z}}{{\partial r}} & = {z_r} = \frac{{{p^2}}}{{{t^3}}} + p\,{{\bf{e}}^{2p + 3r + 4t}} + 3pr\,{{\bf{e}}^{2p + 3r + 4t}}\\ \frac{{\partial z}}{{\partial t}} & = {z_t} = - \frac{{3{p^2}\left( {r + 1} \right)}}{{{t^4}}} + 4pr\,{{\bf{e}}^{2p + 3r + 4t}}\end{align*}}\]

The notation used for the derivative doesn’t matter so we used both here just to make sure we’re familiar with both forms.