For any non-zero real number x, let $f(x) + 2 f(\frac{1}{x}) = 3x$. Then, the sum of all possible values of $x$ for which $f(x) = 3$, is

We have a functional equation: $f(x) + 2f\left(\frac{1}{x}\right) = 3x$ for any non-zero real number $x$.

We need to find all values of $x$ where $f(x) = 3$.

Since this equation works for ANY non-zero value of $x$, we can substitute different values to create a system of equations.

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Equation I: $f(x) + 2f\left(\frac{1}{x}\right) = 3x$ (given)

Equation II: Since the original equation works for any $x$, let's substitute $\frac{1}{x}$ in place of $x$:

$f\left(\frac{1}{x}\right) + 2f\left(\frac{1}{\frac{1}{x}}\right) = 3 \cdot \frac{1}{x}$

$f\left(\frac{1}{x}\right) + 2f(x) = \frac{3}{x}$

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Since we want $f(x) = 3$, let's substitute this into both equations:

From Equation I: $3 + 2f\left(\frac{1}{x}\right) = 3x$

$2f\left(\frac{1}{x}\right) = 3x - 3$

$f\left(\frac{1}{x}\right) = \frac{3x - 3}{2}$ ... (III)

From Equation II: $f\left(\frac{1}{x}\right) + 2(3) = \frac{3}{x}$

$f\left(\frac{1}{x}\right) + 6 = \frac{3}{x}$

$f\left(\frac{1}{x}\right) = \frac{3}{x} - 6$ ... (IV)

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Since both (III) and (IV) equal $f\left(\frac{1}{x}\right)$, we can equate them to find $x$:

$\frac{3x - 3}{2} = \frac{3}{x} - 6$

$\frac{3x - 3}{2} = \frac{3 - 6x}{x}$

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Cross-multiplying: $x(3x - 3) = 2(3 - 6x)$

$3x^2 - 3x = 6 - 12x$

$3x^2 + 9x - 6 = 0$

$x^2 + 3x - 2 = 0$

The solution to this equation gives us $x$ such that $f(x) = 3$

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We need the sum of all possible values of $x$ such that $f(x) = 3$

For any quadratic equation $ax^2 + bx + c = 0$, the sum of roots equals $-\frac{b}{a}$.

In our equation $x^2 + 3x - 2 = 0$:

$a = 1$, $b = 3$, $c = -2$

Sum of roots = $-\frac{3}{1} = -3$

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Therefore, the sum of all possible values of $x$ for which $f(x) = 3$ is $-3$.