For the first derivative, you need to factor the expression $ 4x^3-9x^2+6x-1$. Towards this end, first try guessing a root of $ 4x^3-9x^2+6x-1$. The so called "rational roots theorem' will be useful here. Eventually, you might try $x=1$; and this is indeed a root of $4x^3-9x^2+6x-1$ as substitution will verify.
Now $x=1$ is a root of $ 4x^3-9x^2+6x-1$ if and only if $x-1$ is a factor of $ 4x^3-9x^2+6x-1$. This tells you that $4x^3-9x^2+6x-1$ factors as $ 4x^3-9x^2+6x-1 = (x-1)(ax^2+bx+c) $ for some constants $a$, $b$, and $c$. You can figure those out by doing the long division $ {4x^3-9x^2+6x-1\over x-1}= 4x^2-5x+1. $ So $ 4x^3-9x^2+6x-1=(x-1)(4x^2-5x+1). $
And now you need to factor $4x^2-5x+1$: $4x^2-5x+1=(4x-1)(x-1)$
So f'(x)=(x-1)(x-1)(4x-1).
For the second derivative, you could use the quadratic formula to find its zeroes, or factor it: $ 12x^2-18x+6=6(2x^2-3x+1)=6(2x-1)(x-1) $