Find $f^{(n)}(1)$ where $f(x)={1\over x(2-x)}$. Find $f^{(n)}(1)$ where $f(x)={1\over  x(2-x)}$. 
What I did so far: 
$f(x)=(x(2-x))^{-1}$. $f'(x)=-(x(2-x))^{-2}[2-2x]$ $f''(x)=2(x(2-x))^{-3}[2-2x]^2+2(x(2-x))^{-2}$. It confuses me a lot. I know I have to determine where I have an expression multiplied by $2-2x$ and where not, and what is its form. I would appreciate your help.  

Notice you're asked to find $f^{(n)}(1)$, not $f^{(n)}(x)$.  A series expansion around $a=1$ is a nice way to do this without taking any derivatives.  If you can find a power series representation of the form $f(x) = \sum_{n=0}^\infty c_n (x-a)^n$ with positive radius of convergence, then $c_n = \frac{f^{(n)}(a)}{n!}$.  Therefore $f^{(n)}(a) = c_n n!$.
Following the partial fractions hint we have
\begin{align*}
    \frac{1}{x(2-x)} &= \frac{1}{2} \cdot\frac{1}{x} - \frac{1}{2} \cdot\frac{1}{x-2} \\ 
    &= \frac{1}{2} \cdot \frac{1}{1+(x-1)} + \frac{1}{2}\cdot\frac{1}{1 - (x-1)}
\end{align*}
Now we write the fractions as sums of geometric series.  Since
\begin{align*}
    \frac{1}{1-u} &= \sum_{n=0}^\infty u^n \\
    \frac{1}{1+u} = \frac{1}{1-(-u)} &= \sum_{n=0}^\infty(-u)^n = \sum_{n=0}^\infty(-1)^nu^n \\
\end{align*}
(when $|u|< 1$) we have
\begin{align*}
    \frac{1}{2} \cdot \frac{1}{1+(x-1)} + \frac{1}{2}\cdot\frac{1}{1 - (x-1)}
    &= \frac{1}{2} \sum_{n=0}^\infty (-1)^n(x-1)^n + \frac{1}{2}\sum_{n=0}^\infty (x-1)^n \\
    &= \sum_{n=0}^\infty\left(\frac{1 + (-1)^n}{2}\right) (x-1)^n
\end{align*}
By the initial remark we have
$$
f^{(n)}(1) = n! \left(\frac{1 + (-1)^n}{2}\right)
$$
The factor on the right can be simplified if we know the parity of $n$.  If $n$ is odd $(-1)^n = -1$ and the numerator is zero.  If $n$ is even $(-1)^n =1$, the numerator is $2$, and the quotient is $1$.  Therefore
$$
f^{(n)}(1) = \begin{cases} n! & \text{$n$ is even} \\ 0 & \text{$n$ is odd} \end{cases}
$$
