How do I differentiate the following implicit function $$\sqrt{x^2+y^2} = e^{\sin^{-1} \left( \frac{y}{\sqrt{x^2 + y^2}} \right) } \\ \text{find} \quad \frac{d^2x}{dy^2} \quad \text{i.e. prove} \\ \frac{d^2x}{dy^2} = \frac{2 \left( x^2 + y^2 \right) }{\left( x - y \right)^2}$$
I tried by differentiating directly and making trigonometric substitution
 $x=y \cot a$ which gave parametric equation
 A: First method. Let
$$
F(x,y) = \sqrt{x^2 + y^2} - e^{\arcsin{\frac{y}{\sqrt{x^2 + y^2}}}}.
$$
We know that
$$
\frac{dx}{dy} = -\frac{\partial F/\partial y}{\partial F/\partial x};
$$
so,
$$
\frac{d^2x}{dy^2} = -\frac{d}{dy}\frac{\partial F/\partial y}{\partial F/\partial x}.
$$
It's very long way.
Second method. Let's use polar coordinates:
$$
x=\rho\cos\phi,\enspace y = \rho\sin\phi,
$$
and your equation became
$$
\rho = e^\phi.
$$
It's equation of logarithmic spiral. Now, we need to express $dx/dy$ in terms of $\rho$ & $\phi$.
$$
dx = \cos\phi \,d\rho - \rho\sin\phi\,d\phi\\
dy = \sin\phi \,d\rho + \rho\cos\phi\,d\phi.
$$
Then
$$
\frac{d^2x}{dy^2} = \frac{d}{dy}\frac{dx}{dy} = \frac{\partial}{\partial\rho}\left(\frac{dx}{dy} \right)\cdot\frac{\partial\rho}{\partial y} + \frac{\partial}{\partial\phi}\left(\frac{dx}{dy} \right)\cdot\frac{\partial\phi}{\partial y}
$$
Now we just apply this. First of all
$$
\frac{d\rho}{d\phi} = e^\phi = \rho.
$$
Then
$$
\frac{dx}{dy} = \frac{\cos\phi \,d\rho - \rho\sin\phi\,d\phi}{\sin\phi \,d\rho + \rho\cos\phi\,d\phi} = \frac{\cos\phi\dfrac{d\rho}{d\phi} - \rho\sin\phi}{\sin\phi \dfrac{d\rho}{d\phi} + \rho\cos\phi} = \frac{\rho\cos\phi - \rho\sin\phi}{\rho\sin\phi + \rho\cos\phi} = \frac{\cos\phi - \sin\phi}{\cos\phi+\sin\phi}
$$
Hoora! $\dfrac{\partial}{\partial\rho}\left(\dfrac{dx}{dy} \right) = 0$! At the end we evaluate $\partial\phi/\partial y$:
$$
\frac{\partial\phi}{\partial y} = \frac{1}{\dfrac{\partial y}{\partial\phi}} = \frac{1}{\rho\cos\phi} \left(= \frac1x\right)
$$
Now
$$
\frac{d^2x}{dy^2} = \frac{1}{\rho\cos\phi}\frac{\partial}{\partial\phi}\frac{\cos\phi - \sin\phi}{\cos\phi+\sin\phi} = -\frac{2}{1 + 2\cos\phi\sin\phi}\frac{1}{\rho\cos\phi}
$$
or
$$
\frac{d^2x}{dy^2} = -\frac1x\frac{2}{1 + 2\frac{xy}{\rho^2}} = -\frac2x\frac{x^2+y^2}{x^2+y^2 + 2xy} = -\frac{2}{x}\frac{x^2 + y^2}{(x+y)^2}
$$
