Btukfyl

Let us consider a spring which is having some finite mass. By the help of some external agent the spring has been extended by some distance $x$.
Will the restoring force produced in the spring still be directly proportional to the extension just like any massless spring?

In principle, yes. Generally (i.e in problems) springs are indicated as massless because it makes solving easier in particular situations, but Hooke's Law ($F=-kx$) applies nonetheless.

When spring mass becomes non-negligible, such as determining the acceleration of an attached mass or in the case of spring oscillations, then the mass of the spring has to be considered, and the solution requires integrating over the length of the spring.

The law $F=-kx$ is a very good mathematical approximation of the real springs. But any real spring doesn't obey this law precisely. There are always some differences.

Massless spring is obviously a mathematical abstraction as well. In most circumstances we can treat real springs as massless. But not always, it depends on the problem to be solved.

We can use yet another mathematical abstraction: "ideal spring with mass". But we can't require that it always obeys the law $F=-kx$, because it would be not a spring but something weird.

Imagine that we have extended the spring and let it go. Even our mathematically ideal spring can't instantly get to it's original length parts of the spring have mass and can't move very fast. So there will be a period of time when the spring is still extended, but there is no force applied to it.

However this mathematical abstraction can obey the law $F=-kx$ when there are no movements in the system.

Obviously as soon as we start taking into account realistic features of a spring we will lose the linear Hook law (the force is directly proportional to the extension), which is a good approximation for small extensions.

That said we can try to see if the presence of a mass alters the kinematics of the system: so let's think of adding a mass density to a classic ideal spring.
The conclusion is that when now you extend the spring by applying a force on it, you will have some inertia of the spring which will further oppose a rapid extension.

Let's see how this emerges from a simple model.
We can always write its energy like:$$E=K+V$$ where $E$ is the total energy of the system, $K$ its kinetic energy and $V$ its potential energy.

Focusing on the kinetic term, we know that we can write it as an integral over the mass distribution: $$K=frac{1}{2}int_0^L dl ;lambda; v^2(l,t)$$
where $lambda$ is the mass density per unit of length ( $l$ ) and $v(l)$ is the speed of each point of the spring (here $l$ follows the length of the spring, so indicates a point on the spring). Basically we have written the standard kinetic term $dK=frac{1}{2}dm ;v^2$ for every infinitesimal piece of the string of mass $dm$.

Now let's move to the potential. Does the mass influences the potential energy?
Let's neglect relativistic effects (which we neglected in the extimate of the new kinetic term and in the massless model to begin with); we could still have gravity: having a non-zero mass, the points of the string attract each other by gravitational interaction. This would favor configurations of the spring in which it is slightly compressed (so that all the points are slightly nearer to each other). However since this computation looks messy and since I assume your spring will not have a planetary mass while conserving the original elastic constant, we will neglect this effect too.

Note that I just said that if the spring was to be very heavy its elastic constant and therefore its potential energy would change. This is due to the fact that elastic energy is generated by electric bounds between molecules (or lattice ions in the case of a metallic spring). If you want a spring as heavy as a planet you will need its section to be hundreds of kilometer thick: therefore you will have a lot more microscopic bounds opposing the (giant)spring's torsion and compression.

The bottom line is that electric forces between the microscopic elements of the spring are always way more intense than the gravitational forces between them. So in any physically relevant spring, gravitational "self-interaction" will not play a role.

This was to say that the potential energy $V$ will remain that of the original massless spring: $V=frac{1}{2}k Delta x^2$, where $Delta x$ is the extension of the spring.

Now in order to understand how the Hook law is changed you should write down the equations of motion. If we add a constant external force $F_{ext}$ to the system it will act as a power source (modifying the energy with a term $+Delta x; F_{ext}$).

We can obtain the equations of motion just deriving the energy $E$ with respect to the time:
$$frac{dE}{dt}=dot{E}equiv0=int_0^L!!dl;[lambda;dot{v}(l,t);v(l,t)]+Deltadot{ x}left( kDelta x +F_{ext}right)$$
where we recognize that $Deltadot{x}=v(L,t)$ (if $v(0,t)=0$).

In conclusion, the extension resulting from the application of the external force will be given by:$$kDelta x=-F_{ext}-int_0^L!!dl;lambda;dot{v}(l,t);frac{v(l,t)}{v(L,t)}$$

This means that only in the static limit in which the spring has stopped accelerating the Hook law will be again valid.