If the value of $$\frac{3x\sqrt y + 2y\sqrt x}{3x\sqrt y - 2y\sqrt x} - \frac{3x\sqrt y - 2y\sqrt x}{3x\sqrt y + 2y\sqrt x}$$ is same as that of $$\sqrt x \sqrt y,$$ then which of the following relations between x and y is correct?

Given,  $$\frac{3x\sqrt{y}+2y\sqrt{x}}{3x\sqrt{y}-2y\sqrt{x}}-\frac{3x\sqrt{y}-2y\sqrt{x}}{3x\sqrt{y}+2y\sqrt{x}}=\sqrt{x}\sqrt{y}$$

$$=$$>  $$\frac{\left(3x\sqrt{y}+2y\sqrt{x}\right)^2-\left(3x\sqrt{y}-2y\sqrt{x}\right)^2}{\left(3x\sqrt{y}\right)^2-\left(2y\sqrt{x}\right)^2}=\sqrt{x}\sqrt{y}$$

$$=$$>  $$\frac{9x^2y+4y^2x+12xy\sqrt{x}\sqrt{y}-\left[9x^2y+4y^2x-12xy\sqrt{x}\sqrt{y}\right]}{9x^2y-4y^2x}=\sqrt{x}\sqrt{y}$$

$$=$$>  $$\frac{24xy\sqrt{x}\sqrt{y}}{xy\left(9x-4y\right)}=\sqrt{x}\sqrt{y}$$

$$=$$>  $$\frac{24}{9x-4y}=1$$

$$=$$>  $$9x-4y=24$$

Hence, the correct answer is Option D