It is known that in a non-Abelian group $G$ of order $p^3$, $|Z(G)|=|G'|=p$. In the paper Groups of order $p^6$ (Mathematics of Computation, vol. 34, no. 150) by Rodney James all the groups of orders $p^3, p^4, p^5$ and $p^6$ have been classified. One of the groups of order $p^3$ has the presentation $$G=\langle \alpha, \alpha_{1}, \alpha_{2}| [\alpha_{1}, \alpha]=\alpha_{2}, \alpha^p=\alpha_{1}^p=\alpha_{2}^p=1\rangle.$$
How can we show that for this group $Z(G)=G'=\langle\alpha_{2}\rangle$?
Well...you can't! The presentation does not define a group of order $p^3$. One needs in addition that $[\alpha,\alpha_2]=[\alpha_1,\alpha_2]=1$.