To solve for \(x^3 + 3x\) given that \(x = 7^{\frac{2}{3}} - 7^{-\frac{2}{3}}\), we proceed as follows:

1. First, let's denote \(a = 7^{\frac{1}{3}}\) so that \(a^2 = 7^{\frac{2}{3}}\) and \(a^{-2} = 7^{-\frac{2}{3}}\).
2. This means \(x = a^2 - a^{-2}\).
3. We need to find an expression for \(x^3 + 3x\):
   • Firstly, compute \(x^3\):
     • \(x^3 = (a^2 - a^{-2})^3\)
4. Apply the binomial expansion for \((a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\):
   • Therefore, \((a^2 - a^{-2})^3 = (a^2)^3 - 3(a^2)^2(a^{-2}) + 3(a^2)(a^{-2})^2 - (a^{-2})^3\)
   • Simplify: \((a^6 - 3a^2 + 3a^{-2} - a^{-6})\)
5. Add \(3x\) to \(x^3\):
   • Since \(x = a^2 - a^{-2}\), it follows that \(3x = 3(a^2 - a^{-2}) = 3a^2 - 3a^{-2}\).
   • Adding \(x^3\) and \(3x\), we have:
     • \(x^3 + 3x = (a^6 - 3a^2 + 3a^{-2} - a^{-6}) + (3a^2 - 3a^{-2})\)
     • Combine like terms: \(a^6 - a^{-6}\)
6. Notice that \(a^6 = (7^{\frac{1}{3}})^6 = 7^2 = 49\)
7. Similarly, \(a^{-6} = (7^{-\frac{1}{3}})^6 = 7^{-2} = \frac{1}{49}\).
8. Thus, \(a^6 - a^{-6} = 49 - \frac{1}{49} = \frac{49\times49 - 1}{49} = \frac{2400}{49}\).
9. Therefore, the value of \(x^3 + 3x\) is \(\frac{2400}{49}\).

Hence, the correct answer is

\(\frac{2400}{49}\)