There are a couple of important concepts in this question:
1. Within an instance method, you can access the current object of the same class using 'this'. Therefore, when you access this.myValue, you are accessing the instance member myValue of the ChangeTest instance.
2. If you declare a local variable (or a method parameter) with the same name as the instance field name, the local variable "shadows" the member field. Ideally, you should be able to access the member field in the method directly by using the name of the member (in this example, myValue). However, because of shadowing, when you use myValue, it refers to the local variable instead of the instance field.

In showTwo method, there are two variables accessible with the same name myValue. One is the method parameter and another is the member field of ChangeTest object.
Ideally, you should be able to access the member field in the method directly by using the name myValue but in this case, the method parameter shadows the member field because it has the same name. So by doing this.myValue, you are changing the instance variable myValue by assigning it the value contained in local variable myValue, which is 200. So in the next line when you print ct.myValue, it prints 200.
Now, in the showOne() method also, there are two variables accessible with the same name myValue. One is the method parameter and another is the member field of ChangeTest object. So when you use myValue, you are actually using the method parameter instead of the member field. Therefore, when you do : myValue = myValue; you are actually assigning the value
contained in method parameter myValue to itself. You are not changing the member
field myValue. Hence, when you do System.out.println(ct.myValue); in the next line, it
still prints 200.