Calculate the second derivative of this rational function I'm studying for my exam Math and I came across a problem with one exercise. 

Calculate the second derivative of $$f'(x)=\frac{2x^3-3x^2}{(x^2-1)^2}$$

I just can't seem to calculate the second derivative of this rational function. If someone could help me . 
I just can't seem to get the right answer 
See my calculations and what the right answer should be according to my teacher

Thanks! 

I split the fraction into two parts, then used the product rule on both parts.
$\dfrac{2x^3-3x^2}{(x^2-1)^2} = (2x^3*(x^2-1)^{-2}) - (3x^2*(x^2-1)^{-2})$
After product rule on both terms we have
$\dfrac{12x^3-8x^4}{(x^2-1)^3} + \dfrac{6x^2-6x}{(x^2-1)^2}$
$\dfrac{12x^3-8x^4}{(x^2-1)^3} + \dfrac{(6x^2-6x)*(x^2-1)}{(x^2-1)^{2}*(x^2-1)}$ 
After simplifying the numerators and factoring out a $-2x$, we are left with
$f''(x)=\dfrac{-2x(x^3-3x^2+3x-3)}{(x^2-1)^{3}}$
