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[tex]x^2\dfrac{\mathrm dy}{\mathrm dx}-2xy=4y^4[/tex]
[tex]x^2y^{-4}\dfrac{\mathrm dy}{\mathrm dx}-2xy^{-3}=4[/tex]

Substitute [tex]z=y^{-3}[/tex], so that [tex]\dfrac{\mathrm dz}{\mathrm dx}=-3y^{-4}\dfrac{\mathrm dy}{\mathrm dx}[/tex]. Then the ODE in terms of [tex]z(x)[/tex] is


[tex]-\dfrac13x^2\dfrac{\mathrm dz}{\mathrm dx}-2xz=4[/tex]
[tex]x^2\dfrac{\mathrm dz}{\mathrm dx}+6xz=-12[/tex]
[tex]x^6\dfrac{\mathrm dz}{\mathrm dx}+6x^5z=-12x^4[/tex]

[tex]\dfrac{\mathrm d}{\mathrm dx}\left[x^6z\right]=-12x^4[/tex]
[tex]\implies x^6z=-\dfrac{12}5x^5+C[/tex]
[tex]\implies z=-\dfrac{12}{5x}+\dfrac C{x^6}[/tex]
[tex]\implies\dfrac1{y^3}=-\dfrac{12}{5x}+\dfrac C{x^6}[/tex]

and you can attempt to solve for [tex]y[/tex] explicitly from here if you wish.
In any case, given that [tex]y(1)=\dfrac12[/tex] (presumably that's what you meant to write), we'd get

[tex]\dfrac1{\left(\frac12\right)^3}=-\dfrac{12}{5\cdot1}+\dfrac C{1^6}[/tex]
[tex]8=-\dfrac{12}5+C\implies C=\dfrac{52}5[/tex]

giving a particular solution of

[tex]\dfrac1{y^3}=-\dfrac{12}{5x}+\dfrac{52}{5x^6}[/tex]

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