Note: I am making an assumption here about exactly what is wanted, but I think it the likeliest possibility for a pre-calculus algebra course.
For the order $y\le x$, pretend that $y$ is a constant. Then you get this long division:
$\require{enclose}\begin{align} x^2-\phantom{-x^2}y\phantom{+y^4+y^2}\\ x^2+y\enclose{longdiv}{x^4+\phantom{-x^2y}+\phantom{y^4+}y^4}\\ \underline{x^4+\phantom{-}x^2y\,\phantom{+y^4+y^2}}\\ -x^2y+\phantom{y^4+}y^4\\ \underline{-x^2y-\,\phantom{y^4+}y^2}\\ y^4+y^2 \end{align}$
Thus, $x^4+y^4=\left(x^2+y\right)\left(x^2-y\right)+\left(y^4+y^2\right)$, with quotient $x^2-y$ and remainder $y^4+y^2$.
For the order $x\le y$ you treat $x$ as if it were a constant:
$\require{enclose}\begin{align} y^3-\phantom-x^2y^2+x^4y-\phantom{-y}x^6\;\phantom{+x^8+x^4}\\ y+x^2\enclose{longdiv}{y^4+\phantom{-x^2y^3-x^4y^2+-x^6y}+\phantom{x^8+}x^4}\\ \underline{y^4+\phantom-x^2y^3\;\phantom{-x^4y^2+-x^6y+x^8+x^4}}\\ -x^2y^3\,\phantom{-x^4y^2+-x^6y+x^8+x^4}\\ \underline{-x^2y^3-x^4y^2\phantom{+-x^6y+x^8+x^4}}\\ x^4y^2\phantom{+-x^6y+x^8+x^4}\\ \underline{x^4y^2+\phantom-x^6y\phantom{+x^8+x^4}\;\;}\\ -x^6y+\;\phantom{x^8+}x^4\\ \underline{-x^6y-\;\phantom{x^8+}x^8}\\ x^8+x^4 \end{align}$
That is, $y^4+x^4=\left(y+x^2\right)\left(y^3-x^2y^2+x^4y-x^6\right)+\left(x^8+x^4\right)$, with quotient $y^3-x^2y^2+x^4y-x^6$ and remainder $x^8+x^4$.