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