Why is x == (x = y) not the same as (x = y) == x?

As LouisWasserman said, the expression is evaluated left to right. And java doesn't care what "evaluate" actually does, it only cares about generating a (non volatile, final) value to work with.

//the example values
x = 1;
y = 3;

So to calculate the first output of System.out.println(), the following is done:

x == (x = y)
1 == (x = y)
1 == (x = 3) //assign 3 to x, returns 3
1 == 3
false

and to calculate the second:

(x = y) == x
(x = 3) == x //assign 3 to x, returns 3
3 == x
3 == 3
true

Note that the second value will always evaluate to true, regardless of the initial values of x and y, because you are effectively comparing the assignment of a value to the variable it is assigned to, and a = b and b will, evaluated in that order, always be the same by definition.

== is a binary equality operator.

The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.

^{Java 11 Specification > Evaluation Order > Evaluate Left-Hand Operand First}