smc

$\text{(A)}x+z>y+z⟹x>y$, just subtract $z$ from both sides $\text{(B)}x-z>y-z⟹x>y$, just add $z$ to both sides $\text{(C)}xz>yz⟹x>y\text{ if }x>0$, so that means that our desired answer is $\boxed{xz>yz}$. As a check: $\text{(D)}\frac{x}{z^2}>\frac{y}{z^2}⟹x>y$, we can divide $z^2$ safely and without worry because $z^2>0$. $\text{(E)}x{z}^2>y{z}^2⟹x>y$, similar reasoning as above but instead, multiply $z^2$.