Find the derivative of $\frac{(2x−1)e^{−2x}}{(1−x)^2}$ I need to find the derivative of $$\frac{(2x−1)e^{−2x}}{(1−x)^2}$$

I seems very complex to me so I'm wondering if there is a rule or formula I should be using? I attempted it using the chain rule first for the numerator (since I have $ ( 2 x- 1)$ multiplied by $e^{- 2 x}$ as my numerator) and then my plan was to use this rule: $(\frac{u}{v})′=\frac{vu′-uv′}{v^2}$.

It gets messy and complicated. Could someone please explain how you'd attempt this problem?
If you want to differentiate the product of $3$ functions, the product rule is still available:

$$
(fgh)' = ((fg)h)' = (fg)'h + (fg)h' = (f'g + fg')h + fgh' = f'gh + fg'h + fgh'
$$

(and if you had decided to group (gh) or (fh) at the outset, you can check you'll still get the same outcome)

So, if you take $f = 2x -1 $, $g = e^{-2x}$ and $h = (1-x)^{-2}$, you're set (notice I changed the exponent on $h$ to make this a product rule; you could similarly formulate things in terms of the quotient rule if you prefer)
$$ \dfrac{\mathrm{d}}{\mathrm{d}x}f\left(x\right) =
\dfrac{-2{\cdot}\left(2x-1\right){\cdot}{\mathrm{e}}^{-\left(2x\right)}}{{\left(1-x\right)}^{2}}+\dfrac{2{\cdot}\left(2x-1\right){\cdot}{\mathrm{e}}^{-\left(2x\right)}}{{\left(1-x\right)}^{3}}+\dfrac{2{\mathrm{e}}^{-\left(2x\right)}}{{\left(1-x\right)}^{2}}
$$