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Find $f(x)$ such that,

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

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