Answer
$y=(\frac{x\sin x+ \cos x + C}{x^2}^3$
Work Step by Step
Given:
$$y'+6x^{-1}y=3x^{-1}y^{\frac{2}{3}}\cos x$$
$$x\frac{dy}{dx}+\frac{6}{x}y=\frac{3}{x}\cos xy^{\frac{2}{3}}$$
This equation is in form of Bernoulli equation, where $p(x)=\frac{6}{x}, q(x)=\frac{3 \cos x}{x}$ and $n=\frac{2}{3}$
Dividing both sides by $y^{\frac{2}{3}}$
$$\frac{1}{y^{\frac{2}{3}}}\frac{dy}{dx}+\frac{6}{x}y^{\frac{2}{3}}=\frac{3 \cos x}{x}$$
Let $u=y^{\frac{2}{3}} \rightarrow \frac{du}{dx}=\frac{1}{3y^{\frac{2}{3}}}\frac{dy}{dx}$
Then it becomes:
$$3\frac{du}{dx}+\frac{6u}{x}=\frac{3\cos x}{x}$$
$$\frac{du}{dx}+\frac{2u}{x}=\frac{\cos x}{x}$$
The integrating factor is:
$$I(x)=e^{\int \frac{2}{x}dx}=e^{2\ln (x)}=x^2$$
The equation becomes:
$$\frac{d}{dx}(x^2u)=x\cos x$$
Integrating both sides:
$$x^2u=\int x \cos x$$
Let $u=x$ and $dv=\cos dx$
$\rightarrow du=dx, v=\sin x$
$$\int x\cos x dx=x \sin x- \int \sin x dx=x \sin x + \cos x +C$$
where $C$ is a constant of integration
Then:
$$x^2u=x\sin x+ \cos x + C$$
$$u=\frac{x\sin x+ \cos x + C}{x^2}$$
Subtitute when $u=y^{\frac{1}{3}}$ we have:
$$\rightarrow y^{\frac{1}{3}}=\frac{x\sin x+ \cos x + C}{x^2}$$
$$\rightarrow y=(\frac{x\sin x+ \cos x + C}{x^2}^3$$
