Integrating factor $(3y^2-x) + 2y(y^2-3)y'=0$ is a function 
Show that the differential equation $(3y^2-x) + 2y(y^2-3)y' = 0$ admits an integrating factor which is a function of $(x+y^2)$. Hence solve the equation.

I know how to solve this by using Integrating Factor of an exact equation
But question specifically asking solving using a function. How to approach this question.

$(3y^2-x)+2y(y^2-3)y'=0$
$2y\dfrac{dy}{dx}=\dfrac{x-3y^2}{y^2-3}$
$\dfrac{d(y^2)}{dx}=\dfrac{x-3y^2}{y^2-3}$
$\dfrac{du}{dx}=\dfrac{x-3u}{u-3}$ $(\text{Let}~u=y^2)$
Let $\begin{cases}x=t+a\\u=v+b\end{cases}$ ,
Then $\dfrac{dv}{dt}=\dfrac{t+a-3v-3b}{v+b-3}$
Take $a=9$ and $b=3$ , the ODE becomes
$\dfrac{dv}{dt}=\dfrac{t-3v}{v}$
$\dfrac{dv}{dt}=\dfrac{t}{v}-3$
Let $w=\dfrac{v}{t}$ ,
Then $v=tw$
$\dfrac{dv}{dt}=t\dfrac{dw}{dt}+w$
$\therefore t\dfrac{dw}{dt}+w=\dfrac{1}{w}-3$
$t\dfrac{dw}{dt}=\dfrac{1}{w}-3-w$
$t\dfrac{dw}{dt}=-\dfrac{w^2+3w-1}{w}$
$\dfrac{w}{w^2+3w-1}~dw=-\dfrac{dt}{t}$
$\int\dfrac{w}{w^2+3w-1}~dw=-\int\dfrac{dt}{t}$
$\dfrac{(13+3\sqrt{13})\ln(2w+3+\sqrt{13})}{26}+\dfrac{(13-3\sqrt{13})\ln(2w+3-\sqrt{13})}{26}=-\ln t+c$
$(2w+3+\sqrt{13})^{13+3\sqrt{13}}(2w+3-\sqrt{13})^{13-3\sqrt{13}}t^{26}=C$
$\left(\dfrac{2v}{t}+3+\sqrt{13}\right)^{13+3\sqrt{13}}\left(\dfrac{2v}{t}+3-\sqrt{13}\right)^{13-3\sqrt{13}}t^{26}=C$
$(2v+(3+\sqrt{13})t)^{13+3\sqrt{13}}(2v+(3-\sqrt{13})t)^{13-3\sqrt{13}}=C$
$(2(u-3)+(3+\sqrt{13})(x-9))^{13+3\sqrt{13}}(2(u-3)+(3-\sqrt{13})(x-9))^{13-3\sqrt{13}}=C$
$(2(y^2-3)+(3+\sqrt{13})(x-9))^{13+3\sqrt{13}}(2(y^2-3)+(3-\sqrt{13})(x-9))^{13-3\sqrt{13}}=C$
Which is impossible to solve by considering the integrating factor of the function type $x+y^2$
