Let $y=\frac{1}{2}\sin \theta$ where $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, then \[\frac{1+x^2}{x}=\frac{2(\cos \theta -1)}{\sin \theta}=\frac{-4\sin ^2\frac{\theta}{2}}{2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}=-2\tan \frac{\theta}{2}\implies x^2+2\tan \frac{\theta}{2}x+1=0 \implies x=\frac{-2\tan \frac{\theta}{2}\pm \sqrt{4\tan^2 \frac{\theta}{2}-4}}{2}=-\tan \frac{\theta}{2}\pm \sqrt{\tan^2 \frac{\theta}{2}-1}.\]Let $t=\tan \frac{\theta}{2}$, so \[xy=\frac{1}{1+t^2}\left(-t\pm \sqrt{t^2-1}\right)\]but since we're looking for the maximum value, we can take the positive sign, \[xy=\frac{1}{1+t^2}\left(\sqrt{t^2-1}-t\right)\]now, let $f(t)=\frac{1}{1+t^2}\left(\sqrt{t^2-1}-t\right)$, by differentiating it, we find that the local maximum occurs at $t=-\sqrt{1+\frac{2}{\sqrt{3}}}$, plugging it back yields, \[xy=\frac{\frac{\sqrt 2}{\sqrt[4]{3}}+\sqrt{1+\frac{2}{\sqrt{3}}}}{2+\frac{2}{\sqrt 3}}.\]

Keith50 wrote: Let $y=\frac{1}{2}\sin \theta$ where $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, then \[\frac{1+x^2}{x}=\frac{2(\cos \theta -1)}{\sin \theta}=\frac{-4\sin ^2\frac{\theta}{2}}{2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}=-2\tan \frac{\theta}{2}\implies x^2+2\tan \frac{\theta}{2}x+1=0 \implies x=\frac{-2\tan \frac{\theta}{2}\pm \sqrt{4\tan^2 \frac{\theta}{2}-4}}{2}=-\tan \frac{\theta}{2}\pm \sqrt{\tan^2 \frac{\theta}{2}-1}.\]Let $t=\tan \frac{\theta}{2}$, so \[xy=\frac{1}{1+t^2}\left(-t\pm \sqrt{t^2-1}\right)\]but since we're looking for the maximum value, we can take the positive sign, \[xy=\frac{1}{1+t^2}\left(\sqrt{t^2-1}-t\right)\]now, let $f(t)=\frac{1}{1+t^2}\left(\sqrt{t^2-1}-t\right)$, by differentiating it, we find that the local maximum occurs at $t=-\sqrt{1+\frac{2}{\sqrt{3}}}$, plugging it back yields, \[xy=\frac{\frac{\sqrt 2}{\sqrt[4]{3}}+\sqrt{1+\frac{2}{\sqrt{3}}}}{2+\frac{2}{\sqrt 3}}.\] Are you sure that $y=\frac 1{1+t^2}$ ?