Let's determine all functions $f$ defined on the set of positive real numbers that satisfy the equation $f(x)+2 f\left(\frac{1}{x}\right)=3 x+6$ for all $x$ in their domain.

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Determine all functions $f$ defined on the set of positive real numbers that satisfy the equation $f(x)+2 f\left(\frac{1}{x}\right)=3 x+6$ for all $x$ in their domain.
If in the given equation, $\frac{1}{x}$ is substituted for $x$ everywhere in the set of positive numbers, we get the following relationship:

$$
f\left(\frac{1}{x}\right) + 2 f(x) = 3 \cdot \frac{1}{x} + 6
$$

Together with the original condition, we have a system of two equations with the unknowns $f(x)$ and $f\left(\frac{1}{x}\right)$:

$$
\begin{aligned}
& \quad f(x) + 2 f\left(\frac{1}{x}\right) = 3 x + 6 \\
& (2) \quad f\left(\frac{1}{x}\right) + 2 f(x) = \frac{3}{x} + 6
\end{aligned}
$$

Subtract twice the second equation from the first equation:

$$
-3 f(x) = 3 x - \frac{6}{x} - 6
$$

Thus,

$$
f(x) = \frac{2}{x} - x + 2
$$

Then, $f\left(\frac{1}{x}\right) = 2 x - \frac{1}{x} + 2$, and substituting this back, it can be seen that the obtained function satisfies the condition. Therefore, there is exactly one function that satisfies the equation given in the problem.

Orbán Zsolt (Nagykanizsa, Batthyány Lajos Gymnasium, 8th grade)