Answer
$y=\sqrt \frac{\ln x}{x^2(1-2\ln x)+C}$
Work Step by Step
Given:
$$\frac{dy}{dx}-\frac{1}{2x \ln x}y=2xy^3$$
This equation is in form of Bernoulli equation, where $p(x)=-\frac{1}{2x \ln x}, q(x)=2x$ and $n=3$
Dividing both sides by $y^3$
$$\frac{1}{y^3}\frac{dy}{dx}-\frac{1}{2x \ln x}\frac{1}{y^2}=2x$$
Let $u=y^{-2} \rightarrow \frac{du}{dx}=-2\frac{1}{y^3}\frac{dy}{dx}$
Then it becomes:
$$-\frac{1}{2}\frac{du}{dx}-\frac{u}{2x \ln x}=2x$$
$$\frac{du}{dx}+\frac{u}{x \ln x}=-4x$$
The integrating factor is:
$$I(x)=e^{\int \frac{1}{x \ln x}dx}$$
To solve this, we should let $u=\ln x \rightarrow du=\frac{1}{x}dx \rightarrow \int \frac{1}{x \ln x}dx=\int \frac{1}{u}du = \ln (u)=\ln (\ln (x))$
Hence, the integrating factor is:
$$I(x)=e^{\ln (\ln (x))}=\ln x$$
The equation becomes:
$$\frac{d}{dx}(u \ln x)=\frac{1}{x}u+\ln x\frac{du}{dx}$$
$$\frac{d}{dx}(u \ln x)=-4x \ln x$$
Integrating both sides:
$$\int \frac{d}{dx}(u \ln x)=-4\int x \ln x dx$$
To solve the right side, we let
$u=\ln x, dv=xdx$
$\rightarrow du=\frac{1}{x}dx, v=\frac{x^2}{2}$
$$-4\int x \ln xdx=-4(\frac{x^2}{2}\ln x -\int \frac{x}{2}dx)=-4(\frac{x^2}{2}\ln x-\frac{x^2}{4}+C)$$
where $C$ is a constant of integration
Then:
$$u \ln x =-4(\frac{x^2}{2}\ln x-\frac{x^2}{4}+C)=-2x^2 \ln x+x^2 +C$$
$$u=\frac{x^2}{\ln x}-2x^2+\frac{C}{\ln x}$$
Subtitute when $u=y^{-2}$ we have:
$$\rightarrow y^{-2}=\frac{x^2}{\ln x}-2x^2+\frac{C}{\ln x}$$
$$\rightarrow y=\sqrt \frac{\ln x}{x^2(1-2\ln x)+C}$$
