Question: Integrate $dg = (3x^2/y^2)dx - (2x^3/y^3)dy$ about the path ACB and show your answer in $x_1$, $x_2$, $y_1$, and $y_2$. 
My effort:
I let $dg = 0$ and solve it as an exact differential equation and got $2x^3/y^2 - 2x^3/y = C$. But after that, what should I do?
I know maybe is question is too easy, I apologize for that. Thank you so much for your time!
The meaning of $\;dg = (3x^2/y^2)dx - (2x^3/y^3)dy\;$ is that the right hand is the gradient of a function $\;g\;$ that you have to find. How do we know that? Because
$$\frac{\partial}{\partial x}\left(-\frac{2x^3}{y^3}\right)=\frac{\partial}{\partial y}\left(\frac{3x^2}{y^2}\right)=-\frac{6x^2}{y^3}$$
Thus, find out your function $\;g\;$ . It must fulfill
$$g'_x=\frac{3x^2}{y^2}\stackrel{\text{integration}}\implies g=\frac{x^3}{y^2}+C(y)\;\text{( this last is a constant wrt x)}\;\implies$$
$$g'_y=-\frac{2x^3}{y^3}+C'(y)\stackrel{\text{It must be}}=-\frac{2x^3}{y^3}\implies C'(y)=0\implies C(y)=K=\text{constant}$$
and thus the scalar function $\;g(x,y)=\cfrac{x^3}{y^2}+K\;$ fulfills
$$\nabla g:=\left(g'_x,\,g'_x\right)=\left(\frac{3x^2}{y^2}\,,\;-\frac{2x^3}{y^3}\right)$$
and thus integration is just (Fundamental Theorem of Line Integrals, which is almost the usual FTC)
$$\int_{(x_1,y_1)}^{x_2,y_2)}(3x^2/y^2)dx - (2x^3/y^3)dy=g(x_2,y_2)-g(x_1,y_1)$$
(Observe the constant $\;K\;$ dissapears, as it happens in usual Riemann integral)