Another Year Problem!

Since $3f(x) - 5x * f \left(\dfrac{1}{x} \right) = x - 7$ , (A), it is also the case that

$3f \left( \dfrac{1}{x} \right) - 5 * \dfrac{1}{x} * f \left( \dfrac{1}{\frac{1}{x}} \right) = \dfrac{1}{x} - 7 \Longrightarrow 3f \left( \dfrac{1}{x} \right) - \dfrac{5}{x} f(x) = \dfrac{1}{x} - 7 \Longrightarrow 3x * f \left( \dfrac{1}{x} \right) - 5f(x) = 1 - 7x$ , (B).

Now multiply equation (A) through by 3 and equation (B) through by 5 to find that

$\left( 9f(x) - 15x * f \left( \dfrac{1}{x} \right) \right) + \left( 15x * f \left( \dfrac{1}{x} \right) - 25f(x) \right) = 3(x - 7) + 5(1 - 7x) \Longrightarrow -16f(x) = -32x - 16 \Longrightarrow f(x) = 2x + 1$ .

Finally, $f(2018) = 2 * 2018 + 1 = \boxed{4037}$ .