= Solution
Put $v=y^2$, so $v'=2yy'$. On an <interval> with $x\ne0$, the resulting <first-order linear differential equation> is $v'-4v/x=2x$. Its <integrating factor> $x^{-4}$ gives $(v/x^4)'=2/x^3$, hence
$$
\boxed{y^2=Cx^4-x^2,\qquad y=\pm\sqrt{Cx^4-x^2}.}
$$
For real differentiable graphs one needs $C>0$ and $|x|>1/\sqrt C$, with either fixed sign on a connected <interval>. The points $x=0$ are outside the original equation's domain. At $|x|=1/\sqrt C$, $y=0$ would make the left side zero while the right side is $x\ne0$, so those endpoints are also excluded. The identically zero function is not an additional solution.
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