There have been very similar questions before, but this is what I hope is a fairly comprehensive answer to the question. Don’t be put off by the maths, there’s nothing more complicated going on than a bit of basic linear algebra.
The full expression is:
E[sup]2[/sup] = p[sup]2[/sup]c[sup]2[/sup] + m[sub]0[/sub][sup]2[/sup]c[sup]4[/sup],
where p is momentum and m[sub]0[/sub] is rest mass, or equivalently:
E = γm[sub]0[/sub]c[sup]2[/sup]
where:
γ is the Lorentz factor, which is also the time dilation factor
Note, when the relative velocity of the object to the observer is zero:
E = m[sub]0[/sub]c[sup]2[/sup]
(which is really where the famous E = mc[sup]2[/sup] is useful)
and E goes to infinity as the relative velocity goes to c.
The preference in modern physics is to use the term ‘mass’ for the rest mass m[sub]0[/sub], which all observers agree upon.
How else could mass be defined? (Inertial) mass can be loosely defined as the resistance of a body to a change in motion when a force is applied. In the early days of relativity unsurprisingly physicists were very much married to the Newtonian equation/definition f = ma where f is force and a is acceleration, but in relativity this equation only works as the zero-velocity limit if you use m = m[sub]0[/sub].
You might want to look at what other quantities may work to define mass beyond the zero-velocity limit so that f=ma still works.
In special relativity when a force is applied at right angles to the velocity of an object, the resultant acceleration is given by:
f = γm[sub]0[/sub]a
γm[sub]0[/sub] is sometimes called the “transverse mass” or “relativistic mass”. This does indeed go to infinity as the velocity goes to c. Note also that the momentum is given by:
p = γm[sub]0[/sub]v
Where v is velocity, so on the face of it “relativistic mass” seems like a good concept of mass is special relativity.
However in relativity the direction of the force relative to the direction of motion matters and when a force is applied parallel to the velocity the resultant acceleration is given by:
f = γ[sup]3[/sup]m[sub]0[/sub]a
γ[sup]3[/sup]m[sub]0[/sub] is sometimes called the “longitudinal mass”, which also goes to infinity as velocity goes to c, however for a moving object the longitudinal mass will be larger than the transverse mass.
When you put them both together the relationship between force and acceleration in special relativity becomes:
f = γ[sup]3[/sup]m[sub]0[/sub][(a·v)/v[sup]2[/sup]]v + γm[sub]0[/sub][a - [(a·v)/v[sup]2[/sup]]v]
Note this expression is equivalent to the (longitudinal mass)(the projection of a onto v ) + (transverse mass )(the rejection of a from v).
Now this equation is needlessly complicated a very rarely written down, but it illustrates that there isn’t a scalar in special relativity which you can call “mass = m” to recover f = ma. Or at least that is the case for three dimensional vectors, the relationship between their 4D equivalents four-force F and four-acceleration A is indeed F = m[sub]0[/sub]A. Four-vectors provide a much simpler way to calculate forces and accelerations.
So to answer your question, it does become harder to accelerate an object the closer it is to c, such that you can never accelerate it to c and this can be seen from either the closer the velocity of an object to c, the more kinetic energy you need to add to increase its velocity by 1 unit; or the closer the velocity of an object to c, the less acceleration produced by the same force.
The fact that an object views itself at rest in its own frame doesn’t matter because energy, force and acceleration are also dependent on reference frame.