This can be solved by transformation on the dependent variable using v(y)=y'(x).
It results in a first order ode in v(y). The solution is then found,
then a new first order ode in y(x) now result which is solved again.
Both are separable.
But the integrals are tricky to do . I used Maple for the integrations.
Here is my solution which Maple's verifies. (there are 3 solutions)
https://12000.org/tmp/121121/foo2.pdf
https://12000.org/tmp/121121/foo2.htm
Maple 2021.2 gives these 3 solutions (Maple can't integrate the separable odes)
Intat(1/(-1/2/(1+n)*((3*_a^(1+n)-_C1)*(1+n)^2)^(1/3)-1/2*I*3^(1/2)/(1+n)*((3*_a
^(1+n)-_C1)*(1+n)^2)^(1/3)),_a = y(x))-x-_C2 = 0, Intat(1/(-1/2/(1+n)*((3*_a^(1
+n)-_C1)*(1+n)^2)^(1/3)+1/2*I*3^(1/2)/(1+n)*((3*_a^(1+n)-_C1)*(1+n)^2)^(1/3)),
_a = y(x))-x-_C2 = 0, Intat((1+n)/((3*_a^(1+n)-_C1)*(1+n)^2)^(1/3),_a = y(x))-x
-_C2 = 0
Mathematica 13 can integrate the separable odes' and gives these solutions
in terms of Hypergeometric functions
{{y[x]->InverseFunction[-(((-1/3)^(1/3)*(1+n)^(1/3)*Hypergeometric2F1
[1/3,(1+n)^(-1),1+(1+n)^(-1),-(#1^(1+n)/((1+n)*C[1]))]*#1
*(1+#1^(1+n)/((1+n)*C[1]))^(1/3))/((1+n)*C[1]+#1^(1+n))^(1/3))&]
[x+C[2]]},{y[x]->InverseFunction[((1+n)^(1/3)*Hypergeometric2F1
[1/3,(1+n)^(-1),1+(1+n)^(-1),-(#1^(1+n)/((1+n)*C[1]))]*#1*(1+
#1^(1+n)/((1+n)*C[1]))^(1/3))/(3^(1/3)*((1+n)*C[1]+#1^(1+n))^(1/3))&]
[x+C[2]]},{y[x]->InverseFunction[((-1)^(2/3)*(1+n)^(1/3)
*Hypergeometric2F1[1/3,(1+n)^(-1),1+(1+n)^(-1),-(#1^(1+n)/((1+n)*C[1]))]
*#1*(1+#1^(1+n)/((1+n)*C[1]))^(1/3))/(3^(1/3)*((1+n)*C[1]+#1^(1+n))^(1/3))&][x+C[2]]}}
--Nasser