Let `f''(x)` be continuous at `x=0`
If `lim_(xto0) (2f(x)-3af(2x)+bf(8x))/(sin^(2)x)` exists and `f(0)ne0,f'(0)ne0`, then the value of `3a//b` is________.

Explanation

To solve the problem step-by-step, we will analyze the limit given in the question and derive the necessary equations to find the value of \( \frac{3a}{b} \).

### Step 1: Understanding the Limit
We need to evaluate the limit:
\[ \lim_{x \to 0} \frac{2f(x) - 3af(2x) + bf(8x)}{\sin^2 x} \]
Given that \( f(0) \neq 0 \) and \( f'(0) \neq 0 \), we know that both the numerator and denominator approach 0 as \( x \to 0 \).

### Step 2: Applying L'Hôpital's Rule
Since both the numerator and denominator approach 0, we can apply L'Hôpital's Rule. We differentiate the numerator and denominator.

- The derivative of the numerator:
\[ \frac{d}{dx}[2f(x) - 3af(2x) + bf(8x)] = 2f'(x) - 3a \cdot 2f'(2x) + b \cdot 8f'(8x) \]
This simplifies to:
\[ 2f'(x) - 6af'(2x) + 8bf'(8x) \]

- The derivative of the denominator:
\[ \frac{d}{dx}[\sin^2 x] = 2\sin x \cos x = \sin(2x) \]

### Step 3: Evaluating the Limit Again
Now we need to evaluate:
\[ \lim_{x \to 0} \frac{2f'(x) - 6af'(2x) + 8bf'(8x)}{\sin(2x)} \]
As \( x \to 0 \), \( f'(x) \) approaches \( f'(0) \), and similarly for \( f'(2x) \) and \( f'(8x) \).

### Step 4: Substituting Values
Substituting \( x = 0 \):
\[ \frac{2f'(0) - 6af'(0) + 8bf'(0)}{0} \]
This implies that the numerator must also approach 0 for the limit to exist:
\[ 2f'(0) - 6af'(0) + 8bf'(0) = 0 \]
Factoring out \( f'(0) \) (since \( f'(0) \neq 0 \)):
\[ f'(0)(2 - 6a + 8b) = 0 \]
Thus, we have:
\[ 2 - 6a + 8b = 0 \quad \text{(1)} \]

### Step 5: Finding Another Equation
Returning to the original limit, we can also analyze the first limit we derived:
\[ \lim_{x \to 0} \frac{2f(0) - 3af(0) + bf(0)}{0} = 0 \]
Factoring out \( f(0) \):
\[ f(0)(2 - 3a + b) = 0 \]
Since \( f(0) \neq 0 \):
\[ 2 - 3a + b = 0 \quad \text{(2)} \]

### Step 6: Solving the System of Equations
Now we have two equations:
1. \( 2 - 6a + 8b = 0 \)
2. \( 2 - 3a + b = 0 \)

From equation (2):
\[ b = 3a - 2 \]
Substituting \( b \) into equation (1):
\[ 2 - 6a + 8(3a - 2) = 0 \]
Expanding:
\[ 2 - 6a + 24a - 16 = 0 \]
Combining like terms:
\[ 18a - 14 = 0 \implies 18a = 14 \implies a = \frac{14}{18} = \frac{7}{9} \]

Now substituting \( a \) back to find \( b \):
\[ b = 3\left(\frac{7}{9}\right) - 2 = \frac{21}{9} - \frac{18}{9} = \frac{3}{9} = \frac{1}{3} \]

### Step 7: Finding \( \frac{3a}{b} \)
Now we calculate:
\[ \frac{3a}{b} = \frac{3 \cdot \frac{7}{9}}{\frac{1}{3}} = \frac{\frac{21}{9}}{\frac{1}{3}} = \frac{21}{9} \cdot 3 = \frac{21}{3} = 7 \]

### Final Answer
Thus, the value of \( \frac{3a}{b} \) is:
\[ \boxed{7} \]