Over the functional derivative of a given action

I am trying to get out of this problem, which I know to be basically obvious, but I guess I need a bit of help in: knowing if I am doing the right things, and a bit of mathematical help too.

So I have this action:
$$S = int text{d}t left(dfrac{1}{2}mdot x^2 + lambda x (ddot x)^2right)$$
Where $lambda$ is a constant (real). It is asked to find the equations of motions (Euler-Lagrange) via the functional derivative of the action.

I know that the definition of the FD is

$$delta S[f][h] = dfrac{text{d}}{text{d}epsilon} S[f + epsilon h]bigg|_{epsilon to 0}.$$

But here come the problems, to me. According to the definition I wrote

$$
begin{split}
delta S & = int text{d}t left(dfrac{1}{2}m(dot x(t) + epsilon h(t))^2 + lambdaBig(x(t) + epsilon h(t)Big)Big(ddot x(t) + epsilon ddot h(t) Big)^2right)Bigg|_{epsilon to 0}.
end{split}
$$

Here I start missing a bit the point, for I know somewhere I have to integrate by parts, to reduce the derivative degree, and then with the help of $epsilon to 0$ and the derivative, only few terms will remain.

For example, the first term I compute like this: after having expanded the square, the first term of the square has to be integrated by parts:

$$frac{text{d}}{text{d}epsilon} int text{d}t left( dfrac{1}{2}mdot x(t)^2 + frac{1}{2} mepsilon^2 dot h(t)^2 + mepsilon dot x(t) dot h(t)right)$$

$$frac{text{d}}{text{d}epsilon} left(frac{1}{2}mx(t)dot x(t) – int text{d}t left(frac{1}{2}m x(t)ddot x(t) + frac{1}{2} mepsilon^2 dot h(t)^2 + mepsilon dot x(t) dot h(t)right)right).$$

The first term is zero because there is no $epsilon$, but what about the terms in the integrand?

I cannot mind in a non rigorous way like "the first term in the integrand goes to zero for the same reason, the second one the same because $epsilon^2$ derived leat to another $epsilon$ which vanishes after the limit and the third term is safe", right?

In any case the first term is the simple part. What about the second part? What about the boundary conditions?