We are given the double integral:

$\int_{2}^{6} \int_{0}^{y^2-4} \frac{2x}{3y^5 - 40y^3 + 240y} \, dx \, dy$

Step 1: Evaluate the inner integral with respect to $x$

The integrand has a simple form with respect to $x$:

$\int_0^{y^2 - 4} \frac{2x}{3y^5 - 40y^3 + 240y} \, dx$

The denominator is independent of $x$, so we can treat it as a constant with respect to $x$. Let's first focus on the integral with respect to $x$:

$\int_0^{y^2 - 4} 2x \, dx$

This is a straightforward integral:

$\int 2x \, dx = x^2$

Evaluating this from $0$ to $y^2 - 4$:

$\left[ x^2 \right]_0^{y^2 - 4} = (y^2 - 4)^2 - 0^2 = (y^2 - 4)^2$

Thus, the inner integral simplifies to:

$\frac{(y^2 - 4)^2}{3y^5 - 40y^3 + 240y}$

Step 2: Evaluate the outer integral with respect to $y$

Now, we need to evaluate the outer integral:

$\int_2^6 \frac{(y^2 - 4)^2}{3y^5 - 40y^3 + 240y} \, dy$

First, simplify the numerator:

$(y^2 - 4)^2 = y^4 - 8y^2 + 16$

So the integral becomes:

$\int_2^6 \frac{y^4 - 8y^2 + 16}{3y^5 - 40y^3 + 240y} \, dy$

Now, simplifying the denominator expression:

$3y^5 - 40y^3 + 240y = y(3y^4 - 40y^2 + 240)$

Thus, the integral becomes:

$\int_2^6 \frac{y^4 - 8y^2 + 16}{y(3y^4 - 40y^2 + 240)} \, dy$

This is a more complicated integral, but it can be computed either numerically or by simplifying further. However, due to the structure of the denominator, we might need to use substitution or numeric integration techniques.

Conclusion:

The integral requires either numerical methods or substitution techniques to fully simplify and evaluate. Would you like to proceed with a numeric solution or further simplify this expression?