I have the ODE given below and am told to first find a suitable integrating factor to obtain an implicit solution $F(x,y)=C$ and then solving explicitly for $x$.
$y - 3y^3 = \left(y^4 + 2x\right)y'$ , $y(0)=1$
I shuffled the terms around to get:
$\frac{dy}{dx}+y(\frac{3y^2-1}{y^4+2x})=0$
Then found the integrating factor by doing
$u=e^{^\int\frac{3y^2-1}{y^4+2x}}=e^\frac{(3y^2-1)(ln|y^4+2x|)}{2}$
$e^\frac{(3y^2-1)(ln|y^4+2x|)}{2}y=C$
Then taking $ln$ of both sides and shuffling terms around to finally get:
$x=\frac{e^\frac{2ln|y|}{1-3y^2}-y^4}{2}$
However, this answer is wrong. Is my method wrong?
$$y - 3y^3 = \left(y^4 + 2x\right)\frac{dy}{dx}$$ Your calculus is wrong. You wrote : $$u=e^{\int\frac{3y^2-1}{y^4+2x}}$$ which is non-sens because it is not specified to which variable $x$ , or $y$ , or both related, the integration has to be carried out.
Note that $\int\frac{3y^2-1}{y^4+2x}dx \neq \frac{(3y^2-1)(ln|y^4+2x|)}{2}$ because $y(x)$ is not constant.
A more direct approach is to consider the inverse function $x(y)$ . The ODE becomes : $$(y - 3y^3)x'-2x = y^4 $$ This is a first order linear ODE that you can solve with the standard method.
If you don't see it at first, change of variables : $X=y$ and $Y=x$ which leads to $$(X-3X^3)=(X^4+2Y)\frac{dX}{dY} \quad\to\quad (X-3X^3)\frac{dY}{dX}-2Y=X^4$$ The solution is : $$x=\frac{y^4}{2-6y^2}+C\frac{y^2}{1-3y^2}$$ I am sure that you can take it from here.