Finding Radical of an Ideal 
Given the ideal $J^\prime=\langle xy,xz-yz\rangle$, find it's radical.

I know that the ideal $\langle xy,yz,zx\rangle$ is radical ideal but that's not the case. How can I compute the radical here?

The ideals $J^\prime=\langle xy,xz-yz\rangle\subset J= \langle xy,yz,zx \rangle$ have the same vanishing set in $\mathbb A^3 $, namely the union of the three coordinate axes, so that $\sqrt {J'}=\sqrt J$.
Since you know that $\sqrt J=J$, you can conclude that $$   \sqrt {J'}=J        $$
[The same result can be reached by an easy but uninteresting computation.]  
Edit: answering Danis's questions in his comment
1) Here  or there  are very general result implying  that $J$ is radical.
2) $J'$ is not radical because $xz\notin J'$ but $(xz)^2=xy\cdot z^2+(xz-yz)\cdot xz\in J'$.
