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?

Take $\log$:
\begin{align*}
&\log f(x)\\
&=\log(x-3)+\log(x-2)+\log(x-1)+\log x-\log(x+1)-\log(x+2)-\log(x+3)\\
&\dfrac{f'(x)}{f(x)}\\
&=\dfrac{1}{x-3}+\dfrac{1}{x-2}+\dfrac{1}{x-1}+\dfrac{1}{x}-\dfrac{1}{x+1}-\dfrac{1}{x+2}-\dfrac{1}{x+3}\\
&f'(x)\\
&=f(x)\left(\dfrac{1}{x-3}+\dfrac{1}{x-2}+\dfrac{1}{x-1}+\dfrac{1}{x}-\dfrac{1}{x+1}-\dfrac{1}{x+2}-\dfrac{1}{x+3}\right)
\end{align*}
and then plugging in $x=0$. But first cancel the same factor $x$ in $f(x)\cdot\dfrac{1}{x}$ and then plugging in $x=0$. For the other factors, they are zero.
