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

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

I.F. $=e^{\int \frac{1}{1+y^2}dy}=e^{{\tan }^{-1}y}$

General solution will be

$x\left(e^{{\tan }^{-1}y}\right)=\int e^{{\tan }^{-1}y}\times \frac{{\tan }^{-1}y}{1+y^2}dy$

$\Rightarrow x\left(e^{{\tan }^{-1}y}\right)={\tan }^{-1}y\left(e^{{\tan }^{-1}y}\right)-e^{{\tan }^{-1}y}+C$

Since the curve passes through $\left(1,0\right)$  $\Rightarrow C=2$

So, the curve reduces to $x\left(e^{{\tan }^{-1}y}\right)={\tan }^{-1}y\left(e^{{\tan }^{-1}y}\right)-e^{{\tan }^{-1}y}+2$

when $y=\tan 1, x\left(e\right)=1\left(e\right)-e+2$

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