To solve the problem using the Fundamental Theorem of Calculus (FTC), follow these steps:

1. Given:
   $f''(x) = -\frac{1}{x^2}$, $f'(1) = 3$, and $f(1) = 6$.
   We aim to find $f(x)$.

2. Step 1: Integrate $f''(x)$ to find $f'(x)$.
   $f'(x) = \int f''(x) \, dx = \int -\frac{1}{x^2} \, dx = \frac{1}{x} + C_1$

3. Step 2: Use $f'(1) = 3$ to solve for $C_1$:
   Substitute $x = 1$ into $f'(x) = \frac{1}{x} + C_1$:
   $3 = \frac{1}{1} + C_1 \implies C_1 = 2$ So, $f'(x) = \frac{1}{x} + 2$.

4. Step 3: Integrate $f'(x)$ to find $f(x)$:
   $f(x) = \int f'(x) \, dx = \int \left(\frac{1}{x} + 2\right) dx = \ln|x| + 2x + C_2$

5. Step 4: Use $f(1) = 6$ to solve for $C_2$:
   Substitute $x = 1$ into $f(x) = \ln|x| + 2x + C_2$:
   $6 = \ln(1) + 2(1) + C_2 \implies 6 = 0 + 2 + C_2 \implies C_2 = 4$ So, $f(x) = \ln|x| + 2x + 4$.

Final Answer:
$f(x) = \ln|x| + 2x + 4$

Would you like me to explain further or verify this in more depth?

Related Questions:

1. What is the role of integration constants in this process?
2. How does the Fundamental Theorem of Calculus connect differentiation and integration here?
3. Can the natural logarithm function appear in other integration problems?
4. What happens to the solution if $x < 0$ (consider $\ln|x|$)?
5. How can we confirm $f(x)$ satisfies the given conditions?

Tip:

Always verify the solution by differentiating back to check it satisfies the original equations!