We have, equation of the curve $\mathit{y}\left(1+{\mathit{x}}^2\right)=2-\mathit{x}...\left(\mathit{i}\right)$

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

Since, the given curve passes through $\mathrm{x}-$ axis i.e., $\mathrm{y}=0$.

$\Rightarrow \mathrm{x}=2$
So, the curve passes through the point $\left(2,0\right)$.

$\therefore$ slope of tangent to the curve $=-\frac{1}{5}$
$\therefore$ Equation of tangent of the curve passing through $\left(2,0\right)$ is
$\mathit{y}-0=-\frac{1}{5}\left(\mathit{x}-2\right)$
$\Rightarrow 5\mathit{y}=-\mathit{x}+2$
$\Rightarrow 5\mathit{y}+\mathit{x}=2$