No. Only x might be an lvalue. y is an rvalue.
Assuming y is the name of an object of a type that is assignment
convertible to the type of x, rather than a macro or enumerated
constant, than y in this expression is most certainly an lvalue.
The assignment statement performs "lvalue to value" conversion,
by accessing the value of the object. The result of this conversion
is a pure value, and no longer an lvalue, but that does not mean that
y itself is not an lvalue.
Note also that this used to be called "lvalue to rvalue" conversion,
but the term "rvalue" is no longer used in the current C standard
except for a footnote that this formerly used term has been replaced
by the phrase "value of an expression".
Here's the actual quote from 6.3.2.1 Lvalues, arrays, and function
designators, paragraph 2, of the current C standard:
"Except when it is the operand of the sizeof operator, the unary &
operator, the ++ operator, the -- operator, or the left operand of the
.. operator or an assignment operator, an lvalue that does not have
array type is converted to the value stored in the designated
object (and is no longer an lvalue). If the lvalue has qualified type,
the value has the unqualified version of the type of the lvalue;
otherwise, the value has the type of the lvalue. If the lvalue has an
incomplete type and does not have array type, the behavior is
undefined."