$y^2=x .....\left(i\right)$

Since, $y^2\ge 0\Rightarrow x\ge 0$

$\Rightarrow 2y\frac{dy}{dx}=1$

$\Rightarrow \frac{dy}{dx}=\frac{1}{2y}=m_1$(slope of tangent)

$xy=k .....\left(ii\right)$

$\Rightarrow y+x\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=\frac{-y}{x}=m_2$(slope of tangent)

For two curve to intersect each other at right angle,

$m_1{m}_2=-1$

$\Rightarrow \left(\frac{1}{2y}\right)\left(\frac{-y}{x}\right)=-1$

$\Rightarrow x=\frac{1}{2}$

From $\left(i\right)$, $y=\pm \frac{1}{\sqrt{2}}$

From $\left(ii\right)$, $k=\pm \frac{1}{2\sqrt{2}}$