Find $f(2)$ given that $f$ is a real-valued function that satisfies the equation
$$4f(x)+\left(\frac23\right)(x^2+2)f\left(x-\frac2x\right)=x^3+1.$$
1. Given the functional equation:
   \[
   4f(x) + \left(\frac{2}{3}\right)(x^2 + 2)f\left(x - \frac{2}{x}\right) = x^3 + 1
   \]

2. Substitute \( x = 1 \):
   \[
   4f(1) + \left(\frac{2}{3}\right)(1^2 + 2)f\left(1 - \frac{2}{1}\right) = 1^3 + 1
   \]
   Simplify the equation:
   \[
   4f(1) + \left(\frac{2}{3}\right)(3)f(-1) = 2
   \]
   \[
   4f(1) + 2f(-1) = 2 \quad \text{(Equation 1)}
   \]

3. Substitute \( x = -1 \):
   \[
   4f(-1) + \left(\frac{2}{3}\right)((-1)^2 + 2)f\left(-1 - \frac{2}{-1}\right) = (-1)^3 + 1
   \]
   Simplify the equation:
   \[
   4f(-1) + \left(\frac{2}{3}\right)(3)f(1) = 0
   \]
   \[
   4f(-1) + 2f(1) = 0 \quad \text{(Equation 2)}
   \]

4. Solve the system of linear equations (Equation 1 and Equation 2):
   \[
   \begin{cases}
   4f(1) + 2f(-1) = 2 \\
   4f(-1) + 2f(1) = 0
   \end{cases}
   \]

5. From Equation 2, solve for \( f(-1) \):
   \[
   4f(-1) + 2f(1) = 0
   \]
   \[
   4f(-1) = -2f(1)
   \]
   \[
   f(-1) = -\frac{1}{2}f(1)
   \]

6. Substitute \( f(-1) = -\frac{1}{2}f(1) \) into Equation 1:
   \[
   4f(1) + 2\left(-\frac{1}{2}f(1)\right) = 2
   \]
   \[
   4f(1) - f(1) = 2
   \]
   \[
   3f(1) = 2
   \]
   \[
   f(1) = \frac{2}{3}
   \]

7. Substitute \( f(1) = \frac{2}{3} \) back into the expression for \( f(-1) \):
   \[
   f(-1) = -\frac{1}{2} \left(\frac{2}{3}\right)
   \]
   \[
   f(-1) = -\frac{1}{3}
   \]

8. Substitute \( x = 2 \) into the original equation:
   \[
   4f(2) + \left(\frac{2}{3}\right)(2^2 + 2)f\left(2 - \frac{2}{2}\right) = 2^3 + 1
   \]
   Simplify the equation:
   \[
   4f(2) + \left(\frac{2}{3}\right)(4 + 2)f(1) = 9
   \]
   \[
   4f(2) + \left(\frac{2}{3}\right)(6)f(1) = 9
   \]
   \[
   4f(2) + 4f(1) = 9
   \]
   Substitute \( f(1) = \frac{2}{3} \):
   \[
   4f(2) + 4\left(\frac{2}{3}\right) = 9
   \]
   \[
   4f(2) + \frac{8}{3} = 9
   \]
   \[
   4f(2) = 9 - \frac{8}{3}
   \]
   \[
   4f(2) = \frac{27}{3} - \frac{8}{3}
   \]
   \[
   4f(2) = \frac{19}{3}
   \]
   \[
   f(2) = \frac{19}{12}
   \]

The final answer is \(\boxed{\frac{19}{12}}\).