'this' keyword in 'prototype'

Question

Here is the code

function SuperType(){
    this.property = 'aa';
}

SuperType.prototype.property='aaa'

SuperType.prototype.getSuperValue = function(){
    console.log("this===s : "+(this===s))
    console.log("this===SuperType.prototype : "+(this===SuperType.prototype))
    return this.property;
}



var s=new SuperType();
console.log(s.getSuperValue())

My question is why this in getSuperValue refer to s ? In a normal sense,it should refer to 'SuperType.prototype'.

At first glance, getSuperValue() is called by s ,so this in getSuperValue scope should refer to s. However s.getSuperValue is actually s.__proto__.getSuperValue() that is why I think this in getSuperValue refer to SuperType.prototype

==========================================

I studied all the answers in How does the "this" keyword work? and failed to find anything explain the question

I real reason is s.getSuperValue() <=> Object.getPrototypeOf(s).getSuperValue.call(s) while I wrongly regard it as s.getSuperValue() <=> Object.getPrototypeOf(s).getSuperValue()

To those who concerns ,just put the answer here

===============================================

Got some new understanding of this issue

even though the function reside in prototype

console.log(s.getSuperValue===s.__proto__.getSuperValue) //true

The lookup process is not related to how this is bound

console.log(s.getSuperValue()) // 'aa'
console.log(s.__proto__.getSuperValue()) //'aaa'

this is only determined exclusively by where the function is called

=================== update ================

A more demonstrating example :

function merge(a, b){
    if (a && b) {
        for (var key in b) {
            a[key] = b[key];
        }
    }
    return a;
};
var a={hello:"world"}

var aa={hi:function () {
    console.log(this.hello)
}}
var aaa=merge(a,aa)
aaa.hi()