Solution:


When we assume \(\frac{x}{2}=\frac{y}{3}=k\) , we have x = 2k and y = 3k.

\(\frac {2x}{x+y}\) + \(\frac {3y}{x-y}\) +\(\frac{x^2}{x^2-y^2}\)

= \(\frac {2 * 2k}{2k + 3k}\) + \(\frac {3 * 3k}{2k - 3k}\) + \(\frac{2k^2}{(2k)^2 - (3k)^2}\)

= \(\frac {4k}{5k}\) + \(\frac {9k}{ -k}\) + \(\frac{4k^2}{4k^2 - 9k^2}\)

= \(\frac {4}{5}\) - 9 - \(\frac {4}{5}\) = -9


Therefore, C is the correct answer.

Answer: C