I suspect 4/2 should actually be 4/3, since 4/2 = 2, while 4/3 would make V the volume of a sphere with radius r. But I'll stick with what's given:
[tex]\displaystyle \frac{dV}{dr} = \lim_{h\to0} \frac{2\pi(r+h)^3-2\pi r^3}{h}[/tex]
[tex]\displaystyle \frac{dV}{dr} = 2\pi \lim_{h\to0} \frac{(r^3+3r^2h+3rh^2+h^3)- r^3}{h}[/tex]
[tex]\displaystyle \frac{dV}{dr} = 2\pi \lim_{h\to0} \frac{3r^2h+3rh^2+h^3}{h}[/tex]
[tex]\displaystyle \frac{dV}{dr} = 2\pi \lim_{h\to0} (3r^2+3rh+h^2)[/tex]
[tex]\displaystyle \frac{dV}{dr} = 2\pi \cdot 3r^2 = \boxed{6\pi r^2}[/tex]
In Mathematica, you can check this result via
D[4/2*Pi*r^3, r]
