Ok, let's find the complex potential just to get you going
If $\displaystyle q(x,y)=q_x(x,y)+iq_y(x,y)=3x+(6-3y){\rm i}$, we have
$\displaystyle {\rm div}q=\frac{\partial q_x}{\partial x}(x,y)+\frac{\partial q_y}{\partial y}(x,y)=3-3=0$, so q is a model flow.
Now to find the complex potential $\displaystyle \Phi$, we find its real part as the function $\displaystyle \phi$ to satisfy $\displaystyle q_x=\frac{\partial \phi}{\partial x}(x,y)=3x, \ q_y=\frac{\partial \phi}{\partial y}(x,y)=6-3y$. The first equation gives $\displaystyle \phi=\frac{3x^2}{2}+g(y)$, and substituting into the second one we get $\displaystyle \frac{\partial (\frac{3x^2}{2}+g(y))}{\partial y}(x,y)=6-3y$ or $\displaystyle g'(y)=6-3y$, so $\displaystyle g(y)=6y-\frac{3y^2}{2}$ and $\displaystyle \phi(x,y)=\frac{3x^2}{2}-\frac{3y^2}{2}+6y$.
Now we find the streamline function $\displaystyle \psi$ as the imaginary part of $\displaystyle \Phi$. Since $\displaystyle \psi$ and $\displaystyle \phi$ are conjugate harmonic, we get that $\displaystyle \frac{\partial \phi}{\partial x}=\frac{\partial \psi}{\partial y}, \ \frac{\partial \phi}{\partial y}=-\frac{\partial \psi}{\partial x}$.
The first one gives $\displaystyle \frac{\partial \psi}{\partial y}=3x$ or $\displaystyle \psi=3xy+h(x)$, and by substituting this into the second one $\displaystyle 3y+h'(x)=-6+3y$, from which $\displaystyle h(x)=-6x$ and finally $\displaystyle \psi(x,y)=3xy-6x$.
We obtain the complex potential as $\displaystyle \Phi(x,y)=\phi(x,y)+{\rm i}\psi(x,y)=\phi(x,y)=\frac{3x^2}{2}-\frac{3y^2}{2}+6y+{\rm i}x(3y-6)$.
Now set $\displaystyle \psi=$constant for the equation of streamlines, etc.