Variational derivative of function with respect to its derivative What is
$$
\frac{\delta f(t)}{\delta \dot{f}(t)}~?
$$
Where $\dot{f}(t) = df/dt$.  
 A: The definition of the functional derivative of a functional $I[g]$ is the distribution $\frac{\delta I}{\delta g}(\tau)$ such that 
$$\left\langle \frac{\delta I}{\delta g}, h\right\rangle := \frac{d}{d\alpha}\bigg\rvert_{\alpha=0} I[g+ \alpha h]$$
for every test function $h$. In our case, assuming to deal with functions which suitably vanish before reaching $\pm \infty$,
$$I[g] = \int_{-\infty}^t g(x)dx$$
so that
$$I[\dot{f}]= f(t)$$
as requested. Going on with the procedure
$$\left\langle \frac{\delta I}{\delta g}, h\right\rangle = 
\frac{d}{d\alpha}|_{\alpha=0}  \int_{-\infty}^t(g(\tau)+ \alpha h(\tau)) d\tau = \int_{-\infty}^t h(\tau) dx = \int_{-\infty}^{+\infty} \theta(t-\tau)h(\tau) d\tau$$
where $\theta(\tau)=1$ for $\tau\geq 0$ and $\theta(\tau)=0$ for $\tau<0$
and so 
$$\frac{\delta f(t)}{\delta \dot{f}}(\tau) = \frac{\delta I}{\delta g}(\tau)= \theta(t-\tau)$$
A: The important thing to keep in mind is that a functional derivative is more like a gradient than an ordinary derivative. The reason that this is an important consideration is because, practically, we always specify functions with (possibly infinite) lists of numbers, be they: Taylor series coefficients, continued fraction constants, a list of constant values (approximating with boxcars), a list of points (connect the dots), Fourier series coefficients, or etc. 
The important part of this consideration is that the function's derivative doesn't carry any information about a constant vertical offset. Thus, because any function of the form $f(t) + c$ has the same derivative, $\dot{f}(t)$, the functional derivative in the question will not be defined in the "direction" that corresponds to the degree of freedom defined by $c$.
In equations, let 
$$\begin{align}
g(t) &\equiv \dot{f}(t) \Rightarrow \\
f(t) - f(t_0) & = \int_{t_0}^t g(t') \operatorname{d} t'\end{align}$$
From there:
$$\begin{align}
\frac{\delta f(t)}{ \delta \dot{f}(\tau)} - \frac{\delta f(t_0)}{ \delta \dot{f}(\tau)} & = \frac{\delta \int_{t_0}^t g(t') \operatorname{d}t'}{ \delta g(\tau)} \\
& = \int_{t_0}^t \delta(t' - \tau) \operatorname{d}t' \\
& = \Theta(t-\tau) \, \Theta(\tau - t_0) - \Theta(t_0 - \tau)\, \Theta(\tau - t).
\end{align}$$
This now satisfies:
$$\frac{\partial}{\partial t} \left(\frac{\delta f(t)}{\delta \dot{f}(\tau)}\right) = \delta(t - \tau),$$
as expected. Because $\dot{f}(t)$ doesn't carry any information about the vertical offset of $f(t)$, only differences of the functional derivative, like above, are well defined.
If the space of functions is limited to those that satisfy $\lim_{t\rightarrow -\infty} f(t) = 0$, then we can take $t_0\rightarrow -\infty$ to get the expression from Valter Moretti's answer.
