for $x^2+y^2=a^2$ show that $y''=-(a^2/y^3)$ For $x^2+y^2=a^2$ show that $y''=-(a^2/y^3)$
I got that
$y^2=a^2-x^2$
$y'=-x/y$
$y''=(-1-y'^2)/y$
But then I get stuck.

I would prefer to use parametric equation for these type of problems
$x=a\cos \alpha$
$y=a\sin \alpha$
$\frac{dy}{d\alpha}=a \cos \alpha$
$\frac{dx}{d\alpha}=-a \sin \alpha$
$\frac{dy}{dx}=- \cot \alpha$
$\frac{{d}^2y}{d{x}^2}=-\frac{d}{d\alpha} \cot \alpha \cdot\frac{d\alpha}{dx}$
$\frac{{d}^2y}{d{x}^2}=\frac{ (\operatorname{cosec} \alpha)^2} {-a \sin \alpha}$
$\frac{{d}^2y}{d{x}^2}=\frac{ 1} {-a (\sin \alpha)^3}$
$\frac{{d}^2y}{d{x}^2}=\frac{-a^2} {y^3}$
