I do not get $\displaystyle y= \left(\frac{5}{a}\right)^{\frac{1}{1- a}}x$.
From $\displaystyle 5y- ax^{a- 1}y^a= 0$, $\displaystyle 5y= ax^{a-1}y^a$.
Divide both sides by $\displaystyle y^a$; $\displaystyle 5y^{1-a}= ax^{a- 1}$.
Divide by 5: $\displaystyle y^{1- a}= \frac{a}{5}x^{a- 1}$.
Take the 1- a root: $\displaystyle y= \left(\frac{a}{5}\right)^{\frac{1}{1- a}}x^{\frac{a-1}{1-a}}= \left(\frac{a}{5}\right)^{\frac{1}{1- a}}x^{-1}$.
That is, $\displaystyle x^{-1}$, not $\displaystyle x$.