Prove that f(z)=(2z-1)/(z+2) is analytic on its domain and compute its derivative from 1st principle: f'(z) = lim h-->0 {[f(z+h)-f(z)]/h}?

Use the definition of the derivative (as in Calculus 1).
f'(z) = lim(h→0) [f(z+h) - f(z)]/h
.......= lim(h→0) [(2(z+h)-1)/(z+h+2) - (2z-1)/(z+2)]/h
.......= lim(h→0) [(2z+2h-1)/(z+h+2) - (2z-1)/(z+2)] * (1/h)

Use common denominators:
f'(z) = lim(h→0) [((2z+2h-1)(z+2) - (2z-1)(z+h+2)) / ((z+h+2)(z+2))] * (1/h)
.......= lim(h→0) [5h / ((z+h+2)(z+2))] * (1/h)
.......= lim(h→0) 5/((z+h+2)(z+2))
.......= 5/((z+0+2)(z+2))
.......= 5/(z+2)^2.

I hope this helps!