Differentiate $\frac{x^3}{{(x-1)}^2}$ 
Find $\frac{d}{dx}\frac{x^3}{{(x-1)}^2}$

I start by finding the derivative of the denominator, since I have to use the chain rule. 
Thus, I make $u=x-1$ and $g=u^{-2}$. I find that $u'=1$ and $g'=-2u^{-3}$. I then multiply the two together and substitute $u$ in to get:
$$\frac{d}{dx}(x-1)^{2}=2(x-1)$$
After having found the derivative of the denominator I find the derivative of the numerator, which is $3x^2$. With the two derivatives found I apply the quotient rule, which states that 
$$\frac{d}{dx}(\frac{u(x)}{v(x)})=\frac{v'u-vu'}{v^2}$$
and substitute in the numbers
$$\frac{d}{dx}\frac{x^3}{(x-1)^2}=\frac{3x^2(x-1)^2-2x^3(x-1)}{(x-1)^4}$$
Can I simplify this any further?Is the derivation correct?

No it isn't. When you apply the quotient rule you should be differentiating the denominator $v(x) = (x-1)^2$, not its reciprocal.
