Functional derivative of $\int \left( \frac{df^2 }{d^2 x} \right)^2 dx$

According to page 7 of the PDF document

$$\frac{\delta}{\delta f} \int \left( \frac{df^2 }{d^2 x} \right)^2 dx = \int \frac{df^4}{d^4 x} dx$$

I would like help proving this statement.

Although I can show that

$$\int \left( \frac{df^2 }{d^2 x} \right)^2 dx = \int \frac{df^4}{d^4 x} f dx$$

My attempts at "constructing" the functional derivative of this expression isn't dropping the term $f$. I'm not even sure this is the right way to go about solving the problem.

-

In my understanding, both equations you stated are just wrong. However, the first one may carry some truth: Consider the functional $J(f) = \int (\frac{d^2 f}{dx^2})^2 dx$. Then the functional derivative of $J$ (or first variation) of $J$ is (by integration by parts) $$\delta J(f)(h) = \int 2\frac{d^2f}{dx^2}\frac{d^2h}{dx^2}dx = \int 2\frac{d^4 f}{dx^4} h dx.$$ Hence, one may say (if put in a proper framework of function spaces) that the derivative of $J$ is $$J'(f) = 2\frac{d^4 f}{dx^4}.$$
@Dirk, I don't understand the solution. As far as I can see the E-L for this functional is different. Setting $L \left( x, f, {f}_{x} \right)$ the E-L states that the derivative is ${L}_{f} - \frac{d}{dx} {L}_{{f}_{x}}$. Since ${L}_{f}$ vanishes we're left with ${L}_{{f}_{x}} = 2 {f}_{xx}$ deriving by x again yields $2 {f}_{xxx}$ while you have $2 {f}_{xxxx}$. –  Drazick Dec 14 at 7:44