= Solution
On an interval with $x\ne0$, the zero function is one solution. For a nonzero solution, set $u=y^{-2}$. Dividing the equation by two and differentiating $u$ gives the <linear ordinary differential equation>
$$
u'-\frac2xu=-\frac1{x^3}.
$$
Its <integrating factor> is $x^{-2}$, so $(u/x^2)'=-x^{-5}$ and $u=1/(4x^2)+Cx^2$. Absorbing the factor four into the integration constant gives
$$
\boxed{y(x)=\frac{\pm2x}{\sqrt{1+Kx^4}},\quad\text{or}\quad y(x)=0.}
$$
For real solutions, choose a connected interval avoiding $x=0$ on which $1+Kx^4>0$, with a fixed sign. The special straight solutions $y=\pm2x$ are already included at $K=0$. The separate zero solution would be lost by the division used in the substitution.
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