As we increase the temperature, we know the sharp Fermi surface at zero temperature becomes smeared out at finite temperature $T>0$. (Just think of the Fermi-Dirac distribution, there will be no more a sharp kink when $T>0$.)
Would this smeared-out Fermi surface affect the lab measurement such as using the de Haas–van Alphen effect or the Shubnikov–de Haas effect? How can the Fermi surface be measured precisely at finite $T$?
There is a "thermal broadening" of the peaks in a resistivity - $1/H$ graph, where $H$ is the applied magnetic field. If the sample was at $0K$ the peaks would be like Dirac deltas without being infinite in height, of course. One can still determine the maximum of a sine, or a cosine, even though it isn't very "peaky". For this reason, it is possible to obtain the Fermi surface even at finite temperature, provided that the corresponding effects take place. I.e. provided that the sample is pure enough, at a low temperature enough, and that a strong enough magnetic field is applied, and that either its magnetization or resistivity is being measured as the magnetic field's strength is swept across a large enough range.
As you say, for $T>0$ the occupation of states $f(E)$ will not have a sharp cutoff at $E_F$ but it will fall to zero over an energy interval of ~$k_B T$. The question is then how big is $k_BT$?
Well at room temperature, $k_B T \approx 25$ meV while electronic energies are on the order of eV. So yes, $f(E)$ is blurred at the Fermi energy but it is a small effect at non-extreme temperatures.
