The question is a bit subtle. Without knowing, what $x,y,z$ are, we have to take into account the possibility that e.g. $z=\{x\}$, in which case $\{x\}$ is an element of both $A$ and $B$, in spite of the difference between $\{x\}$ occuring in $A$ and $x\ne\{x\}$ ocurring in $B$. In fact, we can have $A=B$, for example if $x=y$. In concreto, letting $x=y=z=\emptyset$, we find that both $A$ and $B$ are the two-element set $\{\{\emptyset\},\emptyset\}$. But in general (i.e. for "typical" $x,y,z$) we have $A\ne B$. However, a proof of $A\ne B$ would require to exhibit an object that is element of one set and not of the other and this may not be easy without additional assumptions about $x,y,z$. (And as my above remark shows, $x\ne y$ must be among these assumptions). Indeed, if $A=B$ then $x\in A$ (because $x\in B$), hence $x=\{x\}$ or $x=y$ or $x=z$. The first case is impossible (with most set theories); the second case corresponds to my remark above; thus let us assume the third case $x=z$. Similarly, we conclude $y\in B$ and from that $y=\{y\}$ or $y=x$ or $y=z$, from which only the case $y=z$ remains interesting. But if $x=z$ and $y=z$ then we are again arriving at $x=y$. Therefore,
$ \{\{x\},y,z\}=\{x,\{y\},z\}\iff x=y$