ans is 1/2

Transform the first equation using logarithms

Taking natural log of both sides of the first equation: $\ln[(2x)^{\ln 2}] = \ln[(3y)^{\ln 3}]$

$(\ln 2) \cdot \ln(2x) = (\ln 3) \cdot \ln(3y)$

$(\ln 2)[\ln 2 + \ln x] = (\ln 3)[\ln 3 + \ln y]$

$(\ln 2)^2 + (\ln 2)(\ln x) = (\ln 3)^2 + (\ln 3)(\ln y)$ ... (1)

Transform the second equation using logarithms

Taking natural log of both sides of the second equation: $\ln[3^{\ln x}] = \ln[2^{\ln y}]$

$(\ln x)(\ln 3) = (\ln y)(\ln 2)$

Therefore: $\frac{\ln x}{\ln y} = \frac{\ln 2}{\ln 3}$ ... (2)

Use substitution

From equation (2): $\ln x = \frac{\ln 2}{\ln 3} \cdot \ln y$

Substitute this into equation (1): $(\ln 2)^2 + (\ln 2) \cdot \frac{\ln 2}{\ln 3} \cdot \ln y = (\ln 3)^2 + (\ln 3)(\ln y)$

$(\ln 2)^2 + \frac{(\ln 2)^2}{\ln 3} \cdot \ln y = (\ln 3)^2 + (\ln 3)(\ln y)$

Solve for ln y

$(\ln 2)^2 - (\ln 3)^2 = (\ln 3)(\ln y) - \frac{(\ln 2)^2}{\ln 3} \cdot \ln y$

$(\ln 2)^2 - (\ln 3)^2 = \ln y \left[(\ln 3) - \frac{(\ln 2)^2}{\ln 3}\right]$

$(\ln 2)^2 - (\ln 3)^2 = \ln y \left[\frac{(\ln 3)^2 - (\ln 2)^2}{\ln 3}\right]$

$(\ln 2)^2 - (\ln 3)^2 = \ln y \left[\frac{(\ln 3)^2 - (\ln 2)^2}{\ln 3}\right]$

$-[(\ln 3)^2 - (\ln 2)^2] = \ln y \left[\frac{(\ln 3)^2 - (\ln 2)^2}{\ln 3}\right]$

Therefore: $\ln y = -\ln 3$, which gives us $y = \frac{1}{3}$

Find x

From equation (2): $\ln x = \frac{\ln 2}{\ln 3} \cdot \ln y = \frac{\ln 2}{\ln 3} \cdot (-\ln 3) = -\ln 2$

$x = e ^{-\ln 2}$

Therefore: $x = \frac{1}{2}$

The answer is $\boxed{\frac{1}{2}}$ for the x-coordinate.