\(j^{2x} = k^{3y} = (j^4)(k^2)\)

Assume \(j = 2\) to simplify things here.
\(j^{2x} = (j^4)(k^2)\)
\(2^{2x} = (2^4)(k^2)\)

Lets try to make the bases equal,
\(2^{2x} = (2^4)(2^2)\)
\(2^{2x} = 2^6\)
i.e. \(2x = 6\) (when the bases are equal, powers are also equal)
\(x = 3\)

This means, \(j = 2\), \(k = 2\), and \(x = 3\)

Now, \(k^{3y} = (j^4)(k^2)\)
\(2^{3y} = (2^4)(2^2)\)
\(2^{3y} = 2^6\)
i.e. i.e. \(3y = 6\) (when the bases are equal, powers are also equal)
\(y = 2\)

We have \(x = 3\) and \(y = 2\)

Plug \(x = 3\) in the option choices and check which option gives \(y = 2\);

A. \(\frac{3x-6}{2x} = \frac{3(3)-6}{2(3)} = \frac{3}{6} = \frac{1}{2}\)

B. \(\frac{2x-6}{3x} = \frac{2(3)-6}{3(3)} = \frac{0}{9} = 0\)

C. \(\frac{2x}{3x-6} = \frac{2(3)}{3(3)-6} = \frac{6}{3} = 2\)

D. \(\frac{3x}{2x-6} = \frac{3(3)}{2(3)-6} = \frac{9}{0}\)

E. \(\frac{2x}{x-3} = \frac{2(3)}{3-3} = \frac{6}{0}\)

Hence, option C