Let $h$ be the height, r be the base radius and V be the volume of the cylinder which can be inscribed in a sphere of diameter 6 units

from the diagram $r^2+\frac{h^2}{4}=9\Rightarrow r^2=9-\frac{h^2}{4}.$

$V=\pi r^{2}h=\pi h\left(9-\frac{h^2}{4}\right)=9\pi h-\frac{\pi }{4}{h}^3.$

 $\frac{dV}{dh}=9\pi -\frac{3\pi }{4}{h}^2.$

$\therefore \frac{dV}{dh}=0\Rightarrow 9\pi -\frac{3\pi }{4}{h}^2=0\Rightarrow h^2=12\Rightarrow h=2\sqrt{3}.$

$\frac{d^{2}V}{d{h}^2}=-\frac{3}{2}\pi h<0\Rightarrow V$ has maximum when $h=2\sqrt{3}$

$\therefore$ Maximum volume $=9\pi \left(2\sqrt{3}\right)-\frac{\pi }{4}{\left(2\sqrt{3}\right)}^3=18\sqrt{3}\pi -6\sqrt{3}\pi =12\sqrt{3}\pi .$