The slope of normal at any point (x, y), x > 0, y > 0 on the curve...

Given,

Slope of normal $= \dfrac{- dx}{dy} = \dfrac{{x}^{2}}{xy - {x}^{2}{y}^{2} - 1}$

${x}^{2}{y}^{2}dx + dx - xydx = {x}^{2}dy$

${x}^{2}{y}^{2}dx + dx = {x}^{2}dy + xydx$

${x}^{2}{y}^{2}dx + dx = x$\left(xdy + ydx\right)

${x}^{2}{y}^{2}dx + dx = xd \left(xy\right)$

$\dfrac{dx}{x} = \dfrac{dxy}{1 + {x}^{2}{y}^{2}}$

$lnkx = \tan^{- 1}\left(xy\right) ⋯ \left(i\right)</p><p>Curve passes though \left(1, 1\right)</p><p>So,$ In k = $\dfrac{\pi}{4}$ ⇒ k = e$\dfrac{\pi}{4}$

Now from $equation$\left(i\right)$</p><p>We get,$\dfrac{\pi}{4} +$\ln x =$\tan^{- 1}$\left(xy\right)</p><p>$xy = \tan \left(\dfrac{\pi}{4} +ℓnx\right)$</p><p>$xy = \left(\dfrac{1 + \tan \left(ℓnx\right)}{1 - \tan \left(ℓnx\right)}\right) ⋯ \left(ii\right)$</p><p>Put x = e in \left(ii\right)$

We get,$e . y \left(e\right)=\dfrac{1 + \tan 1}{1 - \tan 1}$