Finding an integration factor $x^ay^b$ to solve an ODE 
I have to solve an ODE:$$2(y-3x)dx+x\left(3-\frac{4x}{y} \right) dy=0$$

I am given that I have to use the integrating factor x^a(y^b) where a and b are real numbers in order to turn the problem into a solvable exact ODE. The problem is that I am unsure of how to find these real numbers. From my understanding, an ODE is exact with the integrating factor (I) if the partial derivative of the dx w.r.t y is equal to the partial derivative of the dy w.r.t x (both multiplied with I). I tried doing this but I am unsure of how to proceed with the algebra. So far I have:

If anyone can help with the algebra to find the values for a and b that would be appreciated

Multiply by $y$ the DE:
$$2y^2dx-6xydx+3xydy-4x^2dy=0$$
The integrating factor is $\mu (x,y)=xy$
$$y^3dx^2-6(xy)^2dx+3(xy)^2dy-2x^3dy^2=0$$
$$y^3dx^2-2y^2dx^3+x^2dy^3-2x^3dy^2=0$$
Rearrange some terms:
$$(y^3dx^2+x^2dy^3)-2(y^2dx^3+x^3dy^2)=0$$
$$dx^2y^3-2dx^3y^2=0$$
Integration gives us:
$$x^2y^3-2x^3y^2=K$$

To summarize the integrating factor should be $\mu (x,y)=xy^2$.Then the DE becomes exact.
