The wavefunction of the lowest energy state of hydrogen is $$ \psi(r) = \sqrt{\frac{1}{\pi a_0^3}}e^{-r/a_0}$$ and the probability density is $$\frac{1}{\pi a_0^3}e^{-2r/a_0}$$ where $r$ is the radial distance between electron and nucleus and $a_0$ is the Bohr radius.
To find the most probable value of $r$, do I differentiate w.r.t. $r$ to find the maximum of the probability density function? Or differentiate twice?
The "most probable value" of a density function is supposed to mean the point in which the most mass is located. You are looking for the most probable value of $r$ so you need to translate your probability density, which right now is a density on $\mathbb R^3$ to a density on $\mathbb R_{≥0}$, ie the space of radii.
In this case this is sort of simple, the condition is: $$\int_{r_a}^{r_b} p_r(r)\ dr=\int_{B_{r_b}(0)-B_{r_a}(0)}p(x,y,z)\ dV=\int_{r_a}^{r_b}4\pi r^2 \frac1{\pi a_0^3}e^{-2r/a_0}\ dr$$ And you see how you must include a $4\pi r^2$ term.
So explicitly you need to find the value of $r$ where $$\frac{4r^2}{a_0^3}e^{-2r/a_0}$$ is maximal. Since this function is smooth, you can differentiate it to find the local extrema. If you do this you will find that the maximal value is at $a_0$.