Use the given transformation to evaluate the integral.?

Let x = u/v and y = v <==> u = xy and v = y.

So, the boundary curves of R transform as follows:
y = 2x/3 ==> v = (2u/3)^(1/2)
y = 3x ==> v = (3u)^(1/2)

xy = 2/3 ==> u = 2/3
xy = 3 ==> u = 3.

Moreover, the Jacobian ∂(x, y)/∂(u, v) equals
|1/v -u/v^2|
|..0.........1.| = 1/v.

So, change of variables yields
∫∫R 3xy dx dy
= ∫(u = 2/3 to 3) ∫(v = (2u/3)^(1/2) to (3u)^(1/2)) 3u * (|1/v| dv du)
= ∫(u = 2/3 to 3) ∫(v = (2u/3)^(1/2) to (3u)^(1/2)) (3u/v) dv du
= ∫(u = 2/3 to 3) 3u ln v {for v = (2u/3)^(1/2) to (3u)^(1/2)} du
= ∫(u = 2/3 to 3) 3u * (1/2) [ln(3u) - ln(2u/3)] du
= ∫(u = 2/3 to 3) 3u * (1/2) ln(9/2) du, via quotient rule for logs
= (3/4) ln(9/2) u^2 {for u = 2/3 to 3}
= (77/12) ln(9/2).

I hope this helps!