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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.
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