= Solution
Where $y\ne0$ and $x^2\ne1$, the <separable differential equation> gives
$$
\frac{y\,dy}{2y^2+1}=\frac{x\,dx}{x^2-1},\qquad
\frac14\log(2y^2+1)=\frac12\log|x^2-1|+C.
$$
The requested point selects the <implicit solution> $2y^2+1=(x^2-1)^2$. The two real branches approaching that point are
$$
\boxed{y=\pm\frac{x}{\sqrt2}\sqrt{x^2-2},\qquad x>\sqrt2.}
$$
There is an important <initial value problem> qualification: \b[the differential equation as printed has no classical solution through the specified initial point]. Its right side is undefined at $y=0$, and these branches have an infinite one-sided <derivative> as $x\downarrow\sqrt2$. They satisfy the <differential equation> on $x>\sqrt2$ and extend continuously to the requested point. Multiplying by the vanishing denominator produces an <implicit differential equation> with a larger domain; it does not repair the original <initial value problem>.
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