prove that $\frac{d^2y}{dx^2} = -\frac{a}{y^2}$ for $ x = a\arccos\left(\frac{y-a}{a}\right) - \sqrt{2ay-y^2}$ What is the easiest way to prove that $$\frac{d^2y}{dx^2} = -\frac{a}{y^2}$$ for 
$$ x = a\arccos\left(\frac{y-a}{a}\right) - \sqrt{2ay-y^2}$$ where $a$ is a constant. 

I tried solving it by differentiation on the right side and isolating the derivative, I did that twice to get the second derivative, what I wrong with my approach?
Make the variable substitution $v = \frac1a(y-1)$ to simplify the expression to

$$\frac xa = \arccos v - \sqrt{1-v^2}$$

Then, use $(\arccos v )'=-\frac1{\sqrt{1-v^2}}$ and take the derivatives of both sides to get

$$ \frac1a = -\sqrt{\frac{1-v}{1+v}}\cdot\frac{dv}{dx}$$

Proceed by differentiate again ...
WA says this here $$\text{Solve}\left[\left(\frac{y(x) \sqrt{\frac{2
   a}{y(x)}+c_1}}{c_1}-\frac{2 a \tanh
   ^{-1}\left(\frac{\sqrt{\frac{2
   a}{y(x)}+c_1}}{\sqrt{c_1}}\right)}{c_1{}^{3/2}}\right){}^2=(c_2+x){}^2,y(x)\right]$$