$\displaystyle xy' = yln(xy)$

v=ln(xy)

$\displaystyle dv = dy/y + dx/x$

$\displaystyle dy/y = (ln(xy)dx)/x$

$\displaystyle dv - dx/x = V dx/x$

after cleaning up,

dv/v+1 = dx/x

Integrating that^

ln(v+1) = ln x + C

v+1 = Cx

ln(xy) +1 = Cx

Sorry if it is disorganized, but im new to the forums and format.

Im not sure if i solved it right.

Thanks.