(a) This is a Bernoulli equation:
[tex]\dfrac{dy}{dx} + \dfrac{x}{2(x^2+1)} y = x^3y^5[/tex]
[tex]\implies y^{-5} \dfrac{dy}{dx} + \dfrac{x}{2(x^2+1)} y^{-4} = x^3[/tex]
Substitute [tex]z=y^{-4}[/tex] and [tex]\frac{dz}{dx} = -4y^{-5}\frac{dy}{dx}[/tex] to transform the ODE to
[tex]-\dfrac14 \dfrac{dz}{dx} + \dfrac{x}{2(x^2+1)} z = x^3[/tex]
[tex]\implies \dfrac{dz}{dx} - \dfrac{2x}{x^2+1} z = -4x^3[/tex]
which is now linear in [tex]z[/tex]. Using the integrating factor method, the I.F. is
[tex]\mu = \displaystyle \exp\left(\int -\frac{x}{x^2+1} \, dx\right) = \exp\left(-\frac12 \ln(1+x^2)\right) = \dfrac1{\sqrt{1+x^2}}[/tex]
Distribute [tex]\mu[/tex] on both sides to get a derivative of a product on the left side.
[tex]\dfrac1{\sqrt{1+x^2}} \dfrac{dz}{dx} - \dfrac{2x}{(x^2+1)^{3/2}} z = -\dfrac{4x^3}{\sqrt{1+x^2}}[/tex]
[tex]\implies \dfrac{d}{dx} \left(\dfrac1{\sqrt{1+x^2}} z\right) = -\dfrac{4x^3}{\sqrt{1+x^2}}[/tex]
Integrate both sides (the integral on the right can be done by parts) to get
[tex]\displaystyle \frac1{\sqrt{1+x^2}} z = -4 \int \frac{x^3}{\sqrt{1+x^2}} \, dx = -\frac43 (x^2-2) \sqrt{1+x^2} + C[/tex]
Solve for [tex]z[/tex].
[tex]\displaystyle \frac1{\sqrt{1+x^2}} z = -\frac43 (x^2-2) \sqrt{1+x^2} + C[/tex]
[tex]\implies z = -\dfrac43 (x^2-2) (1+x^2) + C \sqrt{1+x^2}[/tex]
Solve for [tex]y[/tex].
[tex]\dfrac1{y^4} = -\dfrac43 (x^2-2) (1+x^2) + C \sqrt{1+x^2}[/tex]
[tex]\implies \boxed{y^4 = -\dfrac3{4(x^2-2) (1+x^2) + C \sqrt{1+x^2}}}[/tex]
You could go on to solve explicitly for [tex]y[/tex] if you like.
(b) This is also a Bernoulli equation:
[tex]x^2y' + 2xy - y^3 = 0[/tex]
[tex]\implies x^2 y^{-3} y' + 2xy^{-2} = 1[/tex]
Substitute [tex]z=y^{-2}[/tex] and [tex]z' = -2y^{-3}y'[/tex].
[tex]-\dfrac{x^2}2 z' + 2xz = 1[/tex]
[tex]\implies z' - \dfrac4x z = -\dfrac2{x^2}[/tex]
Now repeat the method from (a) to solve for [tex]y[/tex].
[tex]\mu = \exp\left(-\displaystyle \int \frac4x \, dx\right) = \dfrac1{x^4}[/tex]
[tex]\implies \dfrac1{x^4} z' - \dfrac4{x^5} z = -\dfrac2{x^6}[/tex]
[tex]\implies \left(\dfrac1{x^4} z\right)' = -\dfrac2{x^6}[/tex]
[tex]\displaystyle \implies \dfrac1{x^4} z = -2 \int \frac{dx}{x^6}[/tex]
[tex]\implies \dfrac1{x^4} z = \dfrac2{5x^5} + C[/tex]
[tex]\implies z = \dfrac2{5x} + Cx^4[/tex]
[tex]\implies \dfrac1{y^2} = \dfrac2{5x} + Cx^4[/tex]
[tex]\implies \boxed{y^2 = \dfrac{5x}{2 + Cx^5}}[/tex]
