Delta function and derivative of function wrt itself

[Trave11er]

Can someone explain the following profound truth: $$\frac {\partial f(x)}{\partial f(y)} =\delta(x-y)$$

 

[economicsnerd]

The expression $\dfrac{\partial f(x)}{\partial g(y)}$ is really just a short-hand, but there are a handful of rules about which manipulations "usually work" with the short-hand. It turns out that the convention you've named---that $\dfrac{\partial f(x)}{\partial f(y)}$ is $1$ if $x=y$ and $0$ if $x\neq y$---gives us a slightly broader notion of "usually".

 

[arildno]

When you see the Dirac delta function in action, you ought to remind yourself:
"How will this look like when I integrate the expression?"

I haven't seen the result you post before, but do remember the two following results:
$$\int_{-\infty}^{\infty}f(x)\delta{(x-y)}dx=f(y)$$
$$\int_{-\infty}^{\infty}f(y)\delta{(x-y)}dy=f(x)$$
Most likely, your result follows from some clever manipulation of these two basic results.

 

[Trave11er]

Thanks for the replies.
Have found the answer in Parr and Yang's book on Density Functional Theory:
For a functional F=F[f] have
$$\delta F = \int \frac {\delta F} {\delta f(y)} \delta f(y) \,dy$$
In a special case that F=F(f), i.e. F is just some function of f it is required that:
$$\frac {\delta F(f(x))} {\delta f(y)} = \frac {dF}{df} \delta(x-y)$$ in order to have:
$$\delta F = \int \frac {\delta F} {\delta f(y)} \delta f(y) \,dx = \frac {dF} {df} \delta f(x)$$
So that taking F = f, get:
$$\frac {\delta f(x)} {\delta f(y)} = \delta (x-y)$$