Find the second derivative of $y=\sqrt{3x^2-2x+1}$

By indices
$y=(3x^2-2x+1)^\frac{1}{2}$
We use chain rule to evaluate $y$. 
Formula $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$
Let $u=3x^2-2x+1, \frac{du}{dx}=6x-2$
And $u^\frac{1}{2}, \frac{dy}{du}=\frac{1}{2}u^\frac{-1}{2}$. 
Substitute $y,u$ and their derivatives into the chain rule formula.
$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=\frac{1}{2}u^\frac{-1}{2}\times{6x}=3xu^\frac{-1}{2}$

$\frac{dy}{dx}=3x(3x^2-2x+1)^\frac{-1}{2}$.
That is all for the first derivative.
To find the second derivative $\frac{d^2y}{dx^2}$ of $\frac{dy}{dx}=3x(3x^2-2x+1)^\frac{-1}{2}$, we use product rule.
Formula for product rule $\frac{d^2y}{dx^2}=V\frac{du}{dx}+U\frac{dv}{dx}$. 
Let $V=3x, \frac{dv}{dx}=3, U=(3x^2-2x+1)^\frac{-1}{2}, \frac{du}{dx}=\frac{-1}{2}(3x^2-2x+1)^\frac{-3}{2}$. 

$\frac{d^2y}{dx^2}=3x(\frac{-1}{2}(3x^2-2x+1)^\frac{-3}{2})+(3x^2-2x+1)^\frac{-1}{2}.3$
 $=\frac{-3x}{2}(3x^2-2x+1)^\frac{-3}{2}+3(3x^2-2x+1)^\frac{-1}{2}$

Factor out common factors if you like.