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.

Implicit differentiation gives
$$
2x+2yy'=0 \tag{*}
$$
Differentiate again (after removing the common factor $2$):
$$
1+(y')^2+yy''=0 \tag{**}
$$
Now (*) implies $y'=-x/y$, so you can substitute in (**):
$$
1+\frac{x^2}{y^2}+yy''=0
$$
Isolate $y''$ and go on:

 $$y''=-\frac{y^2+x^2}{y^3}$$

