Solve $(x^2y^3+y)dx+(x^2y^3-x)dy=0$ 
$$(x^2y^3+y)dx+(x^2y^3-x)dy=0$$

So:
$$\frac{\partial P}{\partial y}=3x^2y^2+1$$
$$\frac{\partial Q}{\partial x}=2xy^3-1$$
The question is how do I find the integrating factor when neither $$\frac{1}{Q}(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x})$$
is not a function of $x$ alone, nor
$$\frac{1}{P}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})$$
is not a function of $y$ alone.
I guess I have to take a more generic approach then, but maybe there's an easier way that I didn't notice.
 A: Given $P(x,y)dx+Q(x,y)dy=0$, if it has a unique solution (up to arbitrary constants), say $\phi(x,y)=\text{const.}$, then it has infinitely many integrating factors.
To see this, take the differentiation of $\phi(x,y)=c$: 
$$
d\phi=\phi_xdx+\phi_ydy=0
$$
but we also have $Pdx+Qdy=0$, so:
$$
\frac{\phi_x}{P}=\frac{\phi_y}{Q}=\mu(x,y)
$$
therefore $\mu(Pdx+Qdy)=d\phi$ and $\mu$ is the desired integrating factor. 
For any continuous function $F(\phi)$ of $\phi$:
$$
\mu F(\phi)(Pdx+Qdy)=F(\phi)d\phi=0
$$
shows that $\mu F(\phi)$ is also an integrating factor. This shows that there are infinitely many choices of integrating factors.
Unfortunately, there are no general tricks for finding the integrating factor, unless you have already found a particular solution for the original equation.
A: $$(x^2y^3+y)dx+(x^2y^3-x)dy=0 \tag 1$$
Of course this ODE is solvable thanks to numerical calculus.
But it seems that the analytical solution cannot be expressed with the presently available standard functions. Since this is probably an academic exercise, one is drawn to think that there is a typo in the equation.
Supposing that the equation is :
$$(x^3y^2+y)dx+(x^2y^3-x)dy=0 \tag 2$$
again no analytic solution at student level of knowledge.
Supposing that the correct equation is :
$$(x^2y^3+y)dx+(x^3y^2-x)dy=0 \tag 3$$
one can find an integrating factor :
$$\mu(x,y)=\frac{x }{y^3}e^{x^2y^2}$$
Then an exact differential is obtained :
$$\frac{x }{y^3}e^{x^2y^2}\left((x^2y^3+y)dx+(x^3y^2-x)dy\right)=0$$
$$e^{x^2y^2}\left((x^3+\frac{x}{y^2})dx+(\frac{x^4}{y}-\frac{x^2}{y^3})dy\right)=0$$
$$d\left(\frac{x^2}{y^2}e^{x^2y^2} \right)=0$$
The solution of Eq.$(3)$ on the form of implicit equation is :
$$\frac{x^2}{y^2}e^{x^2y^2}=C \tag 4$$
Solving for $y$ requires a special function, the Lambert W function.
$$y(x)=\pm\frac{\sqrt{-W(c\,x^4)}}{x}$$
$c=-1/C$.
As a conclusion, one can suspect that there is a typo in Eq;$(1)$. The correct equation is probably Eq.$(3)$ which solution is $(4)$. Since the wording of the question is suspected of mistake, I don't spent time to edit the calculus in details. 
