Standard equation of ellipse centered at origin is : $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
$\Rightarrow$ $b^{2}x^{2}+a^{2}y^{2}=a^{2}b^{2}$
Differentiating on both sides w.r.t $x$, we get $b^{2}\left ( 2x \right )+a^{2}\left ( 2yy{}' \right )=0$ -------$\left ( 1 \right )$
Differentiating once again on both sides w.r.t $x$, we get $2b^{2}+2a^{2}\left ( yy{}''+\left ( y{}' \right )^{2} \right )=0$
$\therefore$ $b^{2}=-a^{2}\left ( yy{}''+\left ( y{}' \right )^{2} \right )$
Putting the value of $b^{2}$ into equation $\left ( 1 \right )$,
$-a^{2}\left ( yy{}'' +\left ( y' \right )^{2}\right )\left ( 2x \right )+a^{2}\left ( 2yy' \right )=0$
Simplifying and rearranging , we get $xyy''+x\left ( y' \right )^{2}-yy'=0$
Option B is correct.