So this is the correct solution where $U(x)$ means "odd" but personally I only did it with one of them , as in , I didn't use the or-statement to show "when either $y$ is even or $z$ is even" , and I don't see the need for that either , but I am apparently wrong.
Aren't $y$ unknown just like $z$ ? So why do we have to write it twice is the point , when it basically means the same thing?
Your Original Solution with 1 Condition is wrong.
The given New Solution with 2 Conditions is also wrong.
I will give Some Ways to get a Correct Solution , though there is Potential for variations & tweaks.
Statement says EVEN number times "other number" is EVEN. It is not saying "other number" is ODD. It is not saying "other number" is EVEN.
Hence , we can not make the "other number" ODD ( or EVEN ) & we have to make it more general.
Your Original Solution covers the Case $EVEN \times ODD$ , while the New Solution includes the Case $ODD \times EVEN$ , though the Case $EVEN \times EVEN$ is not covered , which might require a new term , making it longer.
More-over , we have 2 "INPUT" numbers , hence , there is no need to use 3 variables.
With that thinking , try this :
$\forall x , y : \lnot U(x) \implies \lnot U(xy) \tag{1}$
In Case , we have to worry whether multiplication is commutative or not , then try this :
$\forall x , y : \lnot U(x) \lor \lnot U(y) \implies \lnot U(xy) \tag{2}$
$\forall x , y : \lnot U(x) \implies \lnot U(xy) \lor \lnot U(yx) \tag{3}$
Finally , we might want to consider the Case $ODD \times ODD$ , though that is not strictly necessary here , since it is not given in the Statement. Covering that Case is very easy :
$\forall x , y : \lnot U(x) \lor \lnot U(y) \iff \lnot U(xy) \tag{4}$