Answer
$${y^4} = 2{x^4} + 1$$
Work Step by Step
$$\eqalign{
& {y^3}\left( {{x^4} + 1} \right)y' - {x^3}\left( {{y^4} + 1} \right) = 0,{\text{ }}y\left( 0 \right) = 1 \cr
& {\text{Add }}{x^3}\left( {{y^4} + 1} \right){\text{ to both sides }} \cr
& {y^3}\left( {{x^4} + 1} \right)y' = {x^3}\left( {{y^4} + 1} \right) \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& {y^3}\left( {{x^4} + 1} \right)\frac{{dy}}{{dx}} = {x^3}\left( {{y^4} + 1} \right) \cr
& {\text{Separate the variables}} \cr
& \frac{{{y^3}}}{{{y^4} + 1}}dy = \frac{{{x^3}}}{{{x^4} + 1}}dx \cr
& {\text{Integrate both sides}} \cr
& \int {\frac{{{y^3}}}{{{y^4} + 1}}} dy = \int {\frac{{{x^3}}}{{{x^4} + 1}}} dx \cr
& \frac{1}{4}\int {\frac{{4{y^3}}}{{{y^4} + 1}}} dy = \frac{1}{4}\int {\frac{{4{x^3}}}{{{x^4} + 1}}} dx \cr
& \frac{1}{4}\ln \left( {{y^4} + 1} \right) = \frac{1}{4}\ln \left( {{x^4} + 1} \right) + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}y\left( 0 \right) = 1 \cr
& \frac{1}{4}\ln \left( {{1^4} + 1} \right) = \frac{1}{4}\ln \left( {{0^4} + 1} \right) + C \cr
& \frac{1}{4}\ln \left( 2 \right) = \frac{1}{4}\ln \left( 1 \right) + C \cr
& C = \frac{1}{4}\ln 2 \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& \frac{1}{4}\ln \left( {{y^4} + 1} \right) = \frac{1}{4}\ln \left( {{x^4} + 1} \right) + \frac{1}{4}\ln 2 \cr
& {\text{Multiply both sides by 4}} \cr
& \ln \left( {{y^4} + 1} \right) = \ln \left( {{x^4} + 1} \right) + \ln 2 \cr
& {e^{\ln \left( {{y^4} + 1} \right)}} = {e^{\ln \left( {{x^4} + 1} \right)}}{e^{\ln 2}} \cr
& {y^4} + 1 = 2\left( {{x^4} + 1} \right) \cr
& {y^4} + 1 = 2{x^4} + 2 \cr
& {y^4} = 2{x^4} + 1 \cr} $$
