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To solve the functional equation (3f(x) + 2f\left(\frac{1}{x}\right) = 4x + 1) for (f(x)), follow these steps:

Step 1: Replace (x) with (\frac{1}{x})

Since the equation holds for all (x \neq 0), substitute (x \to \frac{1}{x}):
[3f\left(\frac{1}{x}\right) + 2f(x) = \frac{4}{x} + 1 \quad \text{(Equation 2)}]

Step 2: Set up a system of equations

Let (A = f(x)) and (B = f\left(\frac{1}{x}\right)). We have:
[3A + 2B = 4x + 1 \quad \text{(Equation 1)}]
[2A + 3B = \frac{4}{x} + 1 \quad \text{(Equation 2)}]

Step 3: Eliminate (B)

Multiply Equation 1 by 3 and Equation 2 by 2:
[9A + 6B = 12x + 3]
[4A + 6B = \frac{8}{x} + 2]

Subtract the second equation from the first:
[5A = 12x - \frac{8}{x} + 1]

Step 4: Solve for (A)

[A = \frac{12x}{5} - \frac{8}{5x} + \frac{1}{5}]

Final Answer:
[f(x) = \boxed{\frac{12x^2 + x - 8}{5x}}]

Or alternatively (expanded form):
[f(x) = \boxed{\frac{12}{5}x - \frac{8}{5x} + \frac{1}{5}}]

Both are correct, but the combined fraction is more compact.

(\boxed{\frac{12x^2 + x - 8}{5x}})

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