Relation between demands of $x, y$ and $z$
Let $\min\{x,z\}=\Omega$, where $P_\Omega=P_x+P_z$. Now the problem becomes $U(y,\Omega)=y\Omega$, which is a standard Cobb-Douglas with degree 2 of homogeneity. Now in this case the choice for each good is:
$y^*=\frac{\alpha_y100}{P_y(\alpha_y+\alpha_\Omega)}\implies y^*=\frac{100}{2P_y}\;\;\;\;\;\;\;$in this case $\alpha_y=\alpha_\Omega=1$
For $\Omega$: $\;\;\;\;\;\Omega^*=\frac{\alpha_\Omega100}{P_\Omega(\alpha_y+\alpha_\Omega)}\implies \Omega^*=\frac{100}{2P_\Omega} \implies \Omega^*=\frac{100}{2(P_x+P_z)}$
Now, since $P_x=P_y=P_z$, let $P_x=P_y=P_z=P$ a general price, therefore substituting in our optimums:
$y^*=\frac{100}{2P}\;\;\;\;\;\Omega^*=\frac{100}{4P}$
Now it's straightforward (since we already know $x^*=z^*$ and as this is the optimum for $\min\{x,z\}$ which is $x$ OR $z$) that $y^*=\frac{100}{2P}&gt;\Omega^*=\frac{100}{4P}$, so this implies that:
$y^*&gt;x^*=z^*\;\;$****
Also I found this document, where this question is number 13.
Hope this helps.
Disclaimer: It would be helpful if other people could assess this approximation, since I hadn't never seen this problem before.