$3 f(x)-2 f\left(\frac{1}{x}\right)=x$...(i)
Replace $x$ with $\frac{1}{x}$ now.
From Eq. (i), we get
$\Rightarrow \quad 3 f\left(\frac{1}{x}\right)-2 f(x)=\frac{1}{x}$...(ii)
$\begin{aligned} & \text { Eq. (i) } \times 2+\text { Eq. (ii) } \times 3 \\ & \qquad\left\{6 f(x)-4 f\left(\frac{1}{x}\right)=2 x\right\} \\ & +\{-6 f(x)+9 f(1 / x)\}=2 x+\frac{3}{x} \\ & 5 f\left(\frac{1}{x}\right)=2 x+\frac{3}{x} \\ & f\left(\frac{1}{x}\right)=\frac{2 x}{5}+\frac{3}{5 x}\end{aligned}$
$\begin{aligned} \therefore \quad 3 f(x) & =x+2 f\left(\frac{1}{x}\right) \\ & =x+2\left(\frac{2 x}{5}+\frac{3}{5 x}\right)\end{aligned}$
$\begin{aligned} 3 f(x) & =x+\frac{4 x}{5}+\frac{6}{5 x} \\ \Rightarrow \quad f(x) & =\frac{x}{3}+\frac{4 x}{15}+\frac{2}{5 x} \\ \therefore \quad f^{\prime}(x) & =\frac{1}{3}+\frac{4}{15}-\frac{2}{5 x^2} \\ f^{\prime}(2) & =\frac{1}{3}+\frac{4}{15}-\frac{2}{5(2)^2} \\ & =\frac{1}{3}+\frac{4}{15}-\frac{1}{10} \\ & =\frac{10+8-3}{30}=\frac{15}{30} \\ f^{\prime}(2) & =\frac{1}{2}\end{aligned}$