Question: I am looking at a datasheet where they show a graph: $V\_{RMS}/\sqrt{Hz}$ as a function of frequency = VSD (voltage spectral density). The question is: how can a spectral density be in RMS? The root mean square is averaged over time or frequencies (Parseval's theorem)

Answer: 
I think you are thinking of Power Spectral Density specifically. This is a common detonation for the spectral density for voltage sources (low source impedance). The actual power across the load will depend on the load applied, which is $V\_{rms}^2/R$, assuming the load is significantly larger than the source impedance.

Once the terms are squared, the units will Watts/Hz, as a power spectral density.

For practical use of this, I do come across this when working with op-amp circuits (for example), and in the course of the design will have interest in knowing the rms voltage (and rms current) noise in the circuit. The noise will depend on the bandwidth, and I would use such a graph as you describe to integrate over the lowest and highest frequency in the bandwidth that my circuit will provide to come up with a total rms voltage quantity that I would then expect to have. The integration needs to be done over the square root of the frequency, as power is proportional to frequency directly, so voltage is proportional to the square root of frequency (Hence the per $\sqrt{Hz}$ in the variable). For instance, if the noise was constant over a range of frequencies at 20 nV/$\sqrt{Hz}$ and I had a bandwidth from 100 Hz to 1KHz, the total rms voltage noise would then be (20 nV)$\sqrt{1000-100}$ = 600 nV rms. 

Adding to the answer given the comments below I show this picture clarifying how each "point" on the frequency axis corresponds to an exponential, not sinusoid waveform in time. This also helps explain clearly the meaning of "negative frequencies", in that they are the exponential terms that are rotating clockwise (negative phase change vs time).

[![enter image description here](https://i.stack.imgur.com/EK4PI.png)](https://i.stack.imgur.com/EK4PI.png)

Euler's Identity is a great example showing this in action:

[![enter image description here](https://i.stack.imgur.com/GpqZh.png)](https://i.stack.imgur.com/GpqZh.png)

Here we notice that the magnitude of each exponential is constant vs time: $$|e^{j\omega t}|=1$$

And therefore the rms of the exponential function is the same as its magnitude. And therefore the rms of an individual point in frequency is the same as its magnitude. However the total power of all the points in frequency will be the sum of the squares. If we want to determine the rms of a waveform with multiple freqeuncy points, this would be accomplished by summing the squares which is total power and then taking the square root which is the rms (Scaling in FFT's aside). 

As a quick sanity check, we are familiar with the rms for a sinusoidal waveform with peak amplitude 1 is $\sqrt{2}/2$. For the two impulses shown in the frequency domain for the cosine each with magnitude 1/2 (also shown from Euler's identity), total power is confirmed to be the sum of the squares of the individual amplitudes: $(\frac{1}{2})^2+(\frac{1}{2})^2 = \frac{1}{2}$, and the square root of this, the rms quantity is confirmed to be $\sqrt{2}/2$. Note in this case if we are relating to voltage terms, the resistance is normalized to be 1.
The power spectral density gives the time-averaged average power as a function of frequency, therefore the voltage spectral density $V(f)$ should give the RMS voltage *of the harmonic component at frequency $f$*. 

(RMS voltage is derived such that power computation formulae are the same as their DC equivalents, so it would make logical sense for the voltage spectral density to be RMS voltage.)