The slope of normal at any point $\left(x,y\right),x>0,y>0$ on the curve $y=y\left(x\right)$ is given by $\frac{x^2}{xy-x^2{y}^2-1}$. If the curve passes through the point $\left(1,1\right)$, then $e·y\left(e\right)$ is equal to

Given,

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

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

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

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

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

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

$\ln kx={\tan }^{-1}\left(xy\right) \cdots \left(\mathrm{i}\right)$

Curve passes though $\left(1,1\right)$

So, In $\mathrm{k}=\frac{\pi }{4}\Rightarrow \mathrm{k}={\mathrm{e}}^{\frac{\pi }{4}}$

Now from equation $\left(\mathrm{i}\right)$ 

We get, $\frac{\pi }{4}+\ln x={\tan }^{-1}\left(xy\right)$

$xy=\tan \left(\frac{\pi }{4}+\ell nx\right)$

$xy=\left(\frac{1+\tan \left(\ell nx\right)}{1-\tan \left(\ell nx\right)}\right) \cdots \left(\mathrm{ii}\right)$

Put $x=e$ in $\left(\mathrm{ii}\right)$

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