$x \to \frac{1}{x}$ 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}}$
