The ODE is linear:
[tex]y'=x^4-\dfrac yx[/tex]
[tex]y'+\dfrac yx=x^4[/tex]
Multiplying both sides by [tex]x[/tex] gives
[tex]xy'+y=x^5[/tex]
Notice that the left side can be condensed as the derivative of a product:
[tex](xy)'=x^5[/tex]
Integrating both sides with respect to [tex]x[/tex] yields
[tex]xy=\dfrac{x^6}6+C[/tex]
[tex]\implies y(x)=\dfrac{x^5}6+\dfrac Cx[/tex]
Since [tex]y(1)=1[/tex],
[tex]1=\dfrac16+C\implies C=\dfrac56[/tex]
so that
[tex]\boxed{y(x)=\dfrac{x^5}6+\dfrac5{6x}}[/tex]
