How is Russell’s Paradox equal to the universal set?

You seem confused about the relationship between Russell's Paradox and the non-existence of the so-called universal set. I will try to clarify without referring to the notion of a class.

If you assume the existence of a universal set $U$ such that $\forall x:x\in U$ and your set theory allows for arbitrary subsets and does not disallow $x\in x$, then you can define set $R$ such that $\forall x:[x\in R\iff x\in U \land x\notin x]$. Note the similarities to the standard presentation of Russell's Paradox.

Applying the definition of $R$ to itself, we have $R\in R \iff R\in U \land R\notin R$. Since, by definition, we must have $R\in U$, this is a contradiction. Thus, the existence of $U$, as defined above, results in a contradiction. Thus, $U$ cannot exist.

Thus, the non-existence of the universal set can be proven (in the set theories described here) by using the same kind of contradiction that arises in Russell's Paradox

There are a few problems with your statements.

1. If there is no universal set, then $\{x\mid x=x\}$ is not a set. This is because this collection is the collection of all sets, since every set is equals to itself.

   In some set theories like $\sf NF$ there is a universal set, but in others like $\sf ZFC$ there is none.

2. $\{x\mid x\notin x\}$ need not be equal to the entire class of sets. It is possible to arrange a situation where we have $x\in x$, but still $x=x$. In some set theories this is outright provable, for example in $\sf NF$ where you have a universal set $U$ then we necessarily have $U\in U$. In such situation, as I wrote, the two classes are not equal.

3. Russell's paradox is not that $\{x\mid x\notin x\}=\{x\mid x=x\}$. It is a paradox that shows that the class $\{x\mid x\notin x\}$ is not a set. It begins by assuming otherwise, then it is a set $A$. If $A\in A$, then by definition $A\notin A$; and if $A\notin A$ then by definition $A\in A$. So we have a contradiction, therefore $A$ is not a set.