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

we have 
$\mathrm{L}=\underset{\mathrm{x}\to 0}{lim} \frac{2\mathrm{f}\left(\mathrm{x}\right)-3\mathrm{af}\left(2\mathrm{x}\right)+\mathrm{bf}\left(8\mathrm{x}\right)}{{\sin }^2\mathrm{x}}$
For the limit to exist, we have
$\begin{array}{l}2\mathrm{f}\left(0\right)-3\mathrm{af}\left(0\right)+\mathrm{bf}\left(0\right)=0\\ \mathrm{or} \hspace{0.25em}3\mathrm{a}-\mathrm{b}=2[\because \mathrm{f}(0)\ne 0, \mathrm{given} ]......\left(1\right)\\ \mathrm{or} \mathrm{L}=\underset{\mathrm{x}\to 0}{lim} \frac{2{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)-6{\mathrm{af}}^{\mathrm{\prime }}\left(2\mathrm{x}\right)+8{\mathrm{bf}}^{\mathrm{\prime }}\left(8\mathrm{x}\right)}{2\mathrm{x}}\\ 2-6a+8b=0\\ 3a-4b=1\\ a=\frac{7}{9}, b=\frac{1}{3}\end{array}$