The normal at the point

Given that the curve $2 y+x^{2}=3$ has a normal passing through point $(1,1)$.

Differentiating both the sides w.r.t. $x$,

$2 \frac{d y}{d x}+2 x=0$

Slope of the tangent $\frac{d y}{d x}=-x$

For $(1,1)$ :

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-1$

Equation of the normal:

$\left(\mathrm{y}-\mathrm{y}_{1}\right)=\frac{-1}{\text { Slope of tangent }}\left(\mathrm{x}-\mathrm{x}_{1}\right)$

$\Rightarrow(\mathrm{y}-1)=\frac{-1}{-1}(\mathrm{x}-1)$

$\Rightarrow y-1=x-1$

$\Rightarrow y-x=0$

$\Rightarrow x-y=0$

Hence, option B is correct.