That function, that made me sleep

It is given that:
${3f(x) - 2f(\frac{1}{x})} = x$

On rearranging the above equation, we can write
$3 f(x) = {x + 2 f(\frac{1}{x})}$
or,
$f(x) = {\frac{{x + 2 f(\frac{1}{x})}}{3}} \rightarrow Eqn. 1$

As we can see, the value of the function is dependent on the value of the argument $x$ , and also on the value of function when the multiplicative inverse of the argument is passed in the function as argument.
So, we will now try to find an equation which will establish the relationship between the value of function when an argument is passed through it, and when its multiplicative inverse is passed through it.

On passing $\frac{1}{x}$ as the argument, we can write,

$f(\frac{1}{x}) = {\frac{{\frac{1}{x} + 2 f(x)}}{3}} \rightarrow Eqn. 2$

Putting the value of $f(\frac{1}{x})$ from $Eqn. 2$ into $Eqn. 1$ , we get,

$f(x) = \frac{x}{3} + (\frac{2}{3}){\frac{(\frac{1}{x} + 2 f(x))}{3}}$
$\Rightarrow f(x) = \frac{x}{3} + (\frac{2}{9})(\frac{1}{x} + 2 f(x))$
$\Rightarrow f(x) = \frac{x}{3} + \frac{2}{9x} + {\frac{4}{9}f(x)}$
$\Rightarrow \frac{5}{9}f(x) = \frac{x}{3} + \frac{2}{9x}$

$\Rightarrow f(x) = \frac{9}{5}[\frac{x}{3} + \frac{2}{9x}]$
$\Rightarrow f(x) = \frac{9}{5}[\frac{9 x^2 + 6}{27x}]$
$\Rightarrow f(x) = \frac{9x^2 + 6}{15x}$

Therefore, function $f(x)$ can be written in simplified form as $\boxed{f(x) = \frac{9x^2 + 6}{15x}}$ .

Problem now simply reduces to an evaluation problem and all we need to do is to put in the different values of arguments at which we want to calculate the value of function. $f(x)$ . After inserting values of arguments given in the options, we find that the correct option is:
$f(4) = \frac{1}{2} 5 f(1)$

$3f({x}) - 2f\left(\dfrac{1}{x}\right) = x$ ----- $\boxed{1}$

By replacing $x$ by $\frac{1}{x}$

$3f({\frac{1}{x}}) - 2f\left(\dfrac{1}{\frac{1}{x}}\right) = \frac{1}{x} \\ => 3f({\frac{1}{x}}) - 2f(x) = \frac{1}{x}$ ----- $\boxed{2}$

So, $3 \times \boxed{1} + 2 \times \boxed{2}$ gives, $9f(x) - 6f(\frac{1}{x}) + 6f(\frac{1}{x}) - 4f(x) = 3x + \frac{2}{x}$

=> $9f(x) - 4f(x) = 3x + \frac{2}{x}$

=> $5f(x) = 3x + \frac{2}{x}$

So, $f(x) = \frac{3x + \frac{2}{x}}{5}$

By substituting, x = 1, x=4 gives you. $f(1) = 1$ and $f(4) = \frac{5}{2}$

So the answer is

$\boxed{f(4) = \frac{1}{2} 5 f(1)}$