To solve this, we need to expand both expressions for small values of \(x^2\) and find the coefficient of \(x^4\).

First, consider \((1 - 2x^2)^{-2}\). Using the binomial expansion for \((1 - u)^{-n}\), we have:

\((1 - 2x^2)^{-2} = 1 + 4x^2 + 12x^4 + \cdots\)

Next, consider \((1 + 6x^2)^{\frac{2}{3}}\). Using the binomial expansion for \((1 + u)^{n}\), we have:

\((1 + 6x^2)^{\frac{2}{3}} = 1 + 4x^2 - 4x^4 + \cdots\)

Subtracting these expansions:

\((1 - 2x^2)^{-2} - (1 + 6x^2)^{\frac{2}{3}} = (1 + 4x^2 + 12x^4) - (1 + 4x^2 - 4x^4)\)

\(= 12x^4 + 4x^4 = 16x^4\)

Thus, \(k = 16\).