We have, $\underset{x\to 0}{\lim }\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}$, putting $x=0$, we get $\frac{0}{0}$ form.

So, Applying L'Hospital rule

 $\underset{x\to 0}{\lim }\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}=\underset{x\to 0}{\lim }\frac{2f\text{'}\left(x\right)-6f\text{'}\left(2x\right)+4f\text{'}\left(4x\right)}{2x}$

again, apply L'Hospital rule,

$\underset{x\to 0}{\lim }\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}=\underset{x\to 0}{\lim }\frac{2f\text{'}\text{'}\left(x\right)-12f\text{'}\text{'}\left(2x\right)+16f\text{'}\text{'}\left(4x\right)}{2}$

Putting $x=0$, we get

$\underset{x\to 0}{\lim }\frac{2f\text{'}\text{'}\left(x\right)-12f\text{'}\text{'}\left(2x\right)+16f\text{'}\text{'}\left(4x\right)}{2}=\frac{6f\text{'}\text{'}\left(0\right)}{2}=12$