We will say that the ODE $$P(x,y)dx+Q(x,y)dy=0$$ where $$P, Q$$ are functions of $$x$$ and $$y$$, it is exact if $$P_y=Q_x$$, where $$P_y$$ indicates the partial derivative of $$P$$ with respect to $$y$$ and $$Q_x$$, the partial derivative of $$Q$$ with respect to $$x$$.
An example of an exact ODE would be:$$$(x^3+y^3) dx+3xy^2dy=0$$$ In effect, calling $$P(x,y)=(x^3+y^3), Q(x,y)=3xy^2$$ we have $$P_y=3y^2=Q_x$$.
It is necessary to observe that not all the ODEs are exact, for example
$$$-y^2\cdot dx +(x^2+xy) \cdot dy=0$$$ This is not exact since calling $$P(x,y)=-y^2, Q(x,y)=x^2+xy$$, we have $$$P_y=-2y\neq Q_x=2x+y$$$.
To solve this type of equations we need to find $$U(x,y)$$ such that $$U_x=P$$ and $$U_y=Q$$ and the solution is given by $$U(xy,y)=C$$, where it $$C$$ is a constant.
To solve this type of equations we will proceed as follows.
We have:$$U_x(x,y)=P(x,y)$$. We integrate on both sides of the equality with respect to $$x$$: $$$\displaystyle \int U_x(x,y) \ dx=\int P(x,y) \ dx \Rightarrow U(x,y)=\int P(x,y) \ dx +h(y)$$$
Therefore we have the function we are looking for, except for the fact that we do not know $$h$$, a function that only depends on $$y$$. To find it, we derive the previous expression with respect to $$y$$: $$$\displaystyle U_y(x,y)= \frac{d}{dy} \int P(x,y) dx + h'(y)$$$ Also, we know that $$U_y=Q$$. Therefore equating terms we obtain a diifferentiable equation (that does not depend on $$x$$, since the ODE is exact) and we can find $$h(y)$$.
Once we know $$h(y)$$ , we add it to the expression of $$U(x,y)$$ which, equated to a constant, is the solution of our ODE.
Let's solve the ODE $$$(x^3+y^3)dx+3xy^2dy=0$$$ which we know to be exact.
We know that we are looking for a function $$U(x,y)$$ so that $$U_x(x,y)=P(x,y)$$. As we have already seen we have: $$$U(x,y)=\int P(x,y) \ dx +h(y)=\int (x^3+y^3)dx+h(y)=\frac{x^4}{4}+y^3x+h(y)$$$ where $$h(y)$$ is a function to be determined that only depends on $$y$$. To find this function, let's impose that $$U$$ is a solution, that is $$U_y(x,y)=Q(x,y)$$.
Deriving with respect to $$y$$ we obtain a solution by equating it to a constant: $$$\left . \begin {array} {r} U_y=3y^2x+h'(y) \\ U_y=Q(x,y)=3xy^2 \end{array}\right\} \Rightarrow h'(y)=0 \Rightarrow h(y)=C$$$ where $$C$$ is a constant that is not so important since we will also have a constant in the final solution.
And so, the solution to the ODE will be: $$$\displaystyle U(x,y)=\frac{x^4}{4}+xy^3=C$$$