How to solve it in simple ways, Find $f'(0)$ $f(x) = \dfrac{\left(x-3\right)\left(x-2\right)\left(x-1\right)x}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}$
Find $f'(0)$
By apply the Quotient Rule $\frac{f'g - g'f}{g^2} $, I can find the answer but it's too long.
Is there any other methods to simplify the steps?

Hint:
Use Partial Fraction Decomposition before differentiation
$$f(x)=\dfrac{x(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}=x+\dfrac A{x+1}+\dfrac B{x+2}+\dfrac C{x+3}$$ 
$$(x+1)(x+2)(x+3)f(x)=?$$
Set $x=-1,-2,-3$ to find $A,B,C$
$$f'(0)=1-\dfrac A{1^2}-\dfrac B{2^2}-\dfrac C{3^2}$$
