Let f"(x) be continuous at x = 0. If $\underset{\mathrm{x}\to 0}{\lim }\frac{2\mathrm{f}\left(\mathrm{x}\right)-3\mathrm{a} \mathrm{f}\left(2\mathrm{x}\right)+\mathrm{bf}\left(8\mathrm{x}\right)}{{\sin }^2\mathrm{x}}$ exists and $\mathrm{f}\left(0\right)\ne 0, \mathrm{f}\text{'}\left(0\right)\ne 0$, then the value of $\frac{3\mathrm{a}}{\mathrm{b}}$  is _____

$\begin{array}{l}\mathrm{We} \mathrm{have}\\ \mathrm{L}=\underset{\mathrm{x}\to 0}{\lim } \frac{2\mathrm{f}\left(\mathrm{x}\right)-3\mathrm{af}\left(2\mathrm{x}\right)+\mathrm{bf}\left(8\mathrm{x}\right)}{{\sin }^2\mathrm{x}}\\ \mathrm{For} \mathrm{the} \mathrm{limit} \mathrm{to} \mathrm{exist}, \mathrm{we} \mathrm{have}\\ 2\mathrm{f}\left(0\right)-3\mathrm{af}\left(0\right)+\mathrm{bf}\left(0\right)=0\\ \mathrm{or} 3\mathrm{a}-\mathrm{b}=2 \left[ \mathrm{f}\left(0\right)\ne 0, \mathrm{given}\right] \left(1\right)\\ \mathrm{or} \mathrm{L}=\underset{\mathrm{x}\to 0}{\lim }\frac{2\mathrm{f}\text{'}\left(\mathrm{x}\right)-6\mathrm{a} \mathrm{f}\text{'}\left(2\mathrm{x}\right)+8\mathrm{b} \mathrm{f}\text{'}\left(8\mathrm{x}\right)}{2\mathrm{x}}\\ \mathrm{For} \mathrm{the} \mathrm{limit} \mathrm{to} \mathrm{exist}, \mathrm{we} \mathrm{have}\\ 2\mathrm{f}\text{'}\left(0\right)-6\mathrm{af}\text{'}\left(0\right)+8\mathrm{bf}\text{'}\left(0\right)=0\\ \Rightarrow 3\mathrm{a}-4\mathrm{b}=1 \left[ \mathrm{f}\text{'}\left(0\right)\ne 0, \mathrm{given}\right] \left(2\right)\\ \mathrm{Solving} \mathrm{equations} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{have} \mathrm{a}=\frac{7}{9} \mathrm{and} \mathrm{b}=\frac{1}{3}.\end{array}$