Step 1:

Given : $y^2 = a(b^2 - x^2)$

Differenting with respect to $x$ on both sides we get

$2yy' = -2ax$

Dividing on both sides by $-2$ we get,

$yy' = -ax$ ---------(1)

Step 2:

Differenting once again on both sides:

Using the information in the tool box we use product rule to differentiate yy'.

Let $y = u$ and $v = y'$. Hence $u' = y'$ and $v' = y''$

Applying this to differentiate equ (1) we get

$yy'' + y'.y' = -a$

$yy'' + (y')^2 = -a$ -------(2)

Step 3:

Dividing equ (2) by (1) we get

$\large\frac{[yy'' + (y')^2 = -a]}{[yy' = -ax]}$

We get $\large\frac{[yy'' + (y')^2]}{ yy' }=\frac{ 1}{x}$

on cross multiplying and rearranging we get

$xyy'' + x(y')^2 -yy'' = 0.$