Verify the given integrating factor of equation. Show that following differential equation admits an integrating factor which is a function of $(x+y^2)$.
$(3y^{2}-x)+2y(y^2 -3x)y^{'}=0$
Approach : Write $y^{'}$ as $dy/dx$. Multiply the equation by $dx$ and  $f(x+y^2)$. Now equation is of the form $Mdx+Ndy=0$.
$f$ is integrating factor if 
$dM/dy = dN/dx$
However, this isn't the case in this equation.
I feel that original equation need to be modified in some manner to get the desired result. However, I am unable to get it.
PS : It is not a homework. This is an exam question. And I was trying it for practice. 
 A: At first we see
$$(3y^2-x)+2y(y^2-3x)y'=0$$
may write of the form $(3y^2-x)+(y^2-3x)(y^2)'=0$ or $(3u-x)dx+(u-3x)du=0$. Now with $M=3u-x$ and $N=u-3x$ we have
$$\dfrac{\partial M}{\partial u}=3~~~,~~~\dfrac{\partial N}{\partial z}=-3$$
which shows not exact.
We know if $I(x,u)$ be an integral factor for non-exact equation $Mdx+Ndu=0$ then for $IMdx+INdu=0$ to be exact 
\begin{eqnarray*}
\frac{\partial(IM)}{\partial u}                      &=& \frac{\partial(IN)}{\partial x}                      \\
\frac{\partial I}{\partial u}M+I\frac{\partial M}{\partial u}  &=& \frac{\partial I}{\partial x}N+I\frac{\partial N}{\partial x}  \\
I(\frac{\partial M}{\partial u}-\frac{\partial N}{\partial x}) &=&\frac{\partial I}{\partial x}N-\frac{\partial I}{\partial u}M   
\end{eqnarray*}
in our equation, we see for artificial variable $z=x+u$,
$$p(z)=\dfrac{\frac{\partial M}{\partial u}-\frac{\partial N}{\partial x}}{N-M}=\dfrac{6}{-2(u+x)}=\dfrac{-3}{z}$$
is a variable of $z$, then our integral factor is
$$I=e^{\int p(z)dz}=\dfrac{1}{z^3}=\dfrac{1}{(u+x)^3}=\color{blue}{\dfrac{1}{(y^2+x)^3}}$$
A: I am unsure of your integrating factor function but by $(x+y^2)$ do you mean if 
$(x+y^2) = u$ then $y^2 = -x + u$ ?
Then if $y^2=-x$ satisfies as a solution to the differential equation, the substitution of $y^2 = -x+u$ will lead you to a general solution to the problem.
Hence for $y^2=-x$ , $2y\frac{dy}{dx}=-1$
Substitute these results back into the differential equation:
$$(3(-x)-x)+2y((-x)-3x)\frac{-1}{2y}=0$$
$$(-4x) - (-4x) = 0 $$
