Hello khaled,
We are given to solve:
$$x\frac{dy}{dx}+y=\frac{1}{y^2}$$
Let's first multiply through by $$\frac{y^2}{x}$$:
$$y^2\frac{dy}{dx}+\frac{y^3}{x}=\frac{1}{x}$$
Next, let's make the substitution:
$$v=y^3\,\therefore\,\frac{dv}{dx}=3y^2\frac{dy}{dx}$$
and we now have:
$$\frac{1}{3}\frac{dv}{dx}+\frac{1}{x}v=\frac{1}{x}$$
Multiply through by 3:
$$\frac{dv}{dx}+\frac{3}{x}v=\frac{3}{x}$$
We now have a linear ODE in $v$. Compute the integrating factor:
$$\mu(x)=e^{3\int\frac{dx}{x}}=x^3$$
And the ODE becomes:
$$x^3\frac{dv}{dx}+3x^2v=3x^2$$
The left side may now be rewritten as:
$$\frac{d}{dx}\left(x^3v \right)=3x^2$$
Integrate with respect to $x$:
$$\int\,d\left(x^3v \right)=3\int x^2\,dx$$
$$x^3v=x^3+C$$
$$v=1+Cx^{-3}$$
Back-substitute for $v$:
$$y^3=1+Cx^{-3}$$
Hence, the explicit solution is:
$$y(x)=\sqrt[3]{1+Cx^{-3}}$$