Fake product rule

$\begin{aligned} \frac d{dx} \left(f(x)g(x)\right) & = f'(x) g'(x) & \small \color{#3D99F6} \text{Given and for }f(x) = x^2 \\ 2xg(x) + x^2g'(x) & = 2xg'(x) \\ 2g & = (2-x)g' \\ \frac {g'}g & = \frac 2{2-x} \\ \int \frac 1g \ dg & = \int \frac 2{2-x} \ dx \\ \ln g & = - 2\ln (2-x) + C & \small \color{#3D99F6} C \text{ is the constant of integration.} \\ \ln g & = \ln \frac 1{(x-2)^2} + C \\ \implies g(x) & = \frac {c_1}{(x-2)^2} \\ g(1) & = \frac {c_1}{(1-2)^2} = 1 & \small \color{#3D99F6} \implies c_1 = 1 \\ \implies g(x) & = \frac 1{(x-2)^2} \\ g'(x) & = - \frac 2{(x-2)^3} \end{aligned}$

$\implies g(3) - g'(3) - \left(f'(x) g'(4)\right) = 1 - (-2) - \left(2(4) \left(-\dfrac 2{2^3}\right) \right) = \boxed{5}$

$\displaystyle \begin{aligned} \frac{d}{dx} \left(f \cdot g\right) & = f' \cdot g' & \small \color{#3D99F6} \text{Definition of product rule} \\ {\color{#3D99F6}{f' \cdot g + f \cdot g'}} & = f' \cdot g' \\ f \cdot g' - f' \cdot g' & = - f' \cdot g \\ g' \left(f - f'\right) & = - f' \cdot g \\ \implies \frac{g'}{g} & = - \frac{f'}{f - f'} \\ \implies \frac{g'}{g} & = \frac{f'}{f' - f} & \small \color{#3D99F6} \text{Integrate both sides} \\ \implies \int \frac{g'}{g} \ dx & = \int \frac{f'}{f' - f} \ dx & \small \color{#3D99F6} \text{Note that } \int \frac{du}{u} = \ln u \\ \implies \ln |g| & = \int \frac{f'}{f' - f} \ dx & \small \color{#3D99F6} \text{Exponentiate both sides} \\ \implies g & = \exp\left(\int \frac{f'}{f' - f} \ dx\right) & \small \color{#3D99F6} \text{Let } f = x^2 \text{ and } f' = 2x \\ \implies g & = \exp\left(\int \frac{2x}{2x - x^2} \ dx \right) \\ \implies g & = \exp\left(2\int \frac{1}{2 - x} \ dx \right) \\ \implies g & = \exp\left(- 2\ln |2 - x| + C\right) \\ \implies g & = |2 - x|^{- 2} \cdot e^C & \small \color{#3D99F6} g(1) = 1 \text{, so } C = 0 \\ \implies g & = |2 - x|^{- 2} \\ \implies g & = \frac{1}{(2 - x)^2} \\ \implies g' & = \frac{2}{(2 - x)^3} \\ \implies f' \cdot g' & = \frac{4x}{(2 - x)^3} \end{aligned}$

$\implies \begin{cases} g(3) = \displaystyle \frac{1}{(2 - 3)^2} = 1 \\ g'(3) = \dfrac{2}{(2 - 3)^3} = - 2 \\ f'(4) \cdot g'(4) = \dfrac{4 \cdot 4}{(2 - 4)^3} = - 2 \end{cases}$

$\implies g(3) - g'(3) - \bigg[f'(4) \cdot g'(4)\bigg] = \boxed{5}$