Given,

$y(x+1)dx-x^{2}dy=0$

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

$\Rightarrow \left(\frac{1}{x}+\frac{1}{x^2}\right)dx=\frac{dy}{y}$

Now integrating both side we get,

${\log }_{e}x-\frac{1}{x}={\log }_{e}y+c$

Now on using $y\left(1\right)=e$ we get, $c=-2$

So, the equation of curve becomes ${\log }_{e}x-\frac{1}{x}={\log }_{e}y-2$

$\Rightarrow y=e^{\ln x}-\frac{1}{x}+2$

Hence, $\underset{x\to 0^{+}}{\lim }{e}^{\ln x-1}-\frac{1}{x}+2$$=e^{-\infty }$$=0$.