Given the question $u_t + 3u_x -2u_y = xyu$. My steps are:

$$\frac{dt}{1} = \frac{dx}{3} = \frac{dy}{-2} = \frac{du}{xyu} \Rightarrow x=C_1+3t, y=C_2-2t$$

$$xydt = \frac{du}{u} \Rightarrow (C_1+3t)(C_2-2t)dt = \frac{du}{u}$$

$$\ln u = C_1C_2 t + \frac{t^2}{2} (3C_2 - 2C_1) - 2t^3 +\phi(C_1, C_2)$$

I know then I need to insert base $e$ into both sides to get $u$, but it would be complex. I wonder is there any simple way to solve this? Or do I have something wrong in above equations so that leads to the complex solution?