I am trying to overload << operator and using friend function. Below code chunk works just fine.
template 
class Mystac         
         friend std::ostream& operator<<(std::ostream& s, Mystack<T> const& d);
declares and befriends a non-template operator<< function. So Mystack<int> would have as its friend a non-template function std::ostream& operator<<(std::ostream& s, Mystack<int> const& d);, etc.
template<class T>
std::ostream& operator<<(std::ostream& s, Mystack<T> const& d)
{
    d.print(s);
    return s;
}
defines an operator<< function template.
The two are not the same. When you write std::cout << intstack;, the overload resolution rules resolve it to the non-template operator<< function you declared, but it isn't defined, so you get a linker error.
There's no way to define a non-template function for every instantiation of a class template outside the class template. You can, however, befriend a specialization of your operator<< function template:
// forward declarations
template <class T>
class Mystack;
template <class T>
std::ostream& operator<<(std::ostream& s, Mystack<T> const& d);
template <class T>
class Mystack
{
    friend std::ostream& operator<< <T>(std::ostream& s, Mystack<T> const& d); 
//                                  ^^^
};
or befriend every specialization of the function template, which is worse from an encapsulation point of view (since, e.g., operator<< <int> would be a friend of Mystack<float>):
template <class T>
class Mystack
{
public:
    template <class U>
    friend std::ostream& operator<<(std::ostream& s, Mystack<U> const& d);
};
or just define the friend function inside the class.