Differentiate the given solution:
[tex]y=\dfrac1{x^2+C}\implies y'=-\dfrac{2x}{(x^2+C)^2}[/tex]
Substitute [tex]y[/tex] and [tex]y'[/tex] into the ODE:
[tex]-\dfrac{2x}{(x^2+C)^2}+2x\left(\dfrac1{x^2+C}\right)^2=0[/tex]
and it's easy to see the left side indeed reduces to 0.
