Solution:
$f(-2)= \dfrac{67}{20}$
Step by step solution:
$\quad 3f\left(\dfrac1x\right) + \dfrac{2f(x)}{x}= x^2$
$+)\quad$ With $x = -2$, we get:
$3f\left(-\dfrac12\right) - f(-2)= 4$
It's equal to:
$f\left(-\dfrac12\right)= \dfrac{4 + f(-2)}{3}$
$+)\quad$ With $x = -\dfrac12$, we get:
$3f(-2) - 4f\left(-\dfrac12\right)= \dfrac14$
Then:
$3f(-2) - 4\cdot \dfrac{4+ f(-2)}{3}= \dfrac14$
Or:
$\dfrac{5f(-2)}{3} = \dfrac{67}{12}$
So that:
$f(-2)= \dfrac{67}{20}$
Answer: $f(-2)= \dfrac{67}{20}$