$x \to \frac{1}{x}$ in the original equation yields,

$f\left( {\frac{1}{x}} \right) + f(x) + 3f\left( { – \frac{1}{x}} \right) = \frac{1}{x}$ ……..(A)

Original equation – (A) yields,

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

$x \to – x$ above gives us $3f(x) – 3f\left( {\frac{1}{x}} \right) = – x + \frac{1}{x}$

$ \Rightarrow f(x) – f\left( {\frac{1}{x}} \right) = – \frac{x}{3} + \frac{1}{{3x}}$ ……..(B)

Original equation + (B) gives us $2f(x) + 3f( – x) = \frac{{2x}}{3} + \frac{1}{{3x}}$ ……..(C)

$x \to – x$ above yields,

$2f( – x) + 3f(x) = – \frac{{2x}}{3} – \frac{1}{{3x}}$ ……..(D)

Now, $(D) \times 3 – (C) \times 2$

$5f(x) = – 2x – \frac{1}{x} – \frac{{4x}}{3} – \frac{2}{{3x}} = – \frac{{10x}}{3} – \frac{5}{{3x}}$

$ \Rightarrow f(x) = – \frac{{2x}}{3} – \frac{1}{{3x}}$