If `lim_(xtooo) f(x)` exists and is finite and nonzero and if `lim_(xtooo) {f(x)+(3f(x)-1)/(f^(2)(x))}=3`, then the value of `lim_(xtooo) f(x)" is "`_______.

Explanation

To solve the problem, we will follow these steps:

1. **Given Information**:
We are given that \( \lim_{x \to \infty} f(x) \) exists, is finite, and non-zero. We also know that:
\[ \lim_{x \to \infty} \left( f(x) + \frac{3f(x) - 1}{f^2(x)} \right) = 3 \]

2. **Let \( L = \lim_{x \to \infty} f(x) \)**:
We will denote \( L \) as the limit of \( f(x) \) as \( x \) approaches infinity. Thus, we can rewrite the limit expression:
\[ L + \frac{3L - 1}{L^2} = 3 \]

3. **Rearranging the Equation**:
We can rearrange the equation:
\[ \frac{3L - 1}{L^2} = 3 - L \]
Multiplying both sides by \( L^2 \) (since \( L \neq 0 \)):
\[ 3L - 1 = (3 - L)L^2 \]

4. **Expanding the Right Side**:
Expanding the right side gives:
\[ 3L - 1 = 3L^2 - L^3 \]

5. **Rearranging to Form a Polynomial**:
Rearranging the equation leads to:
\[ L^3 - 3L^2 + 3L - 1 = 0 \]

6. **Factoring the Polynomial**:
We can factor the polynomial:
\[ (L - 1)^3 = 0 \]
This implies:
\[ L - 1 = 0 \quad \Rightarrow \quad L = 1 \]

7. **Conclusion**:
Therefore, we find that:
\[ \lim_{x \to \infty} f(x) = 1 \]

### Final Answer:
The value of \( \lim_{x \to \infty} f(x) \) is \( \boxed{1} \).