Determine the first and second orders derivatives of $y = f(x)$ determined implicity by
$$ \ln\sqrt{x^2 + y^2} = \alpha~\arctan\frac{y}{x} $$
Now, notice that in polar coordinates, the expression for this equation becomes simply
$$ \ln (r) = \alpha \theta $$
Which is much easier to deal with, especially when calculating a second order derivative, since higher order derivatives tend to become "hairy".
So I want to use polar coordinates to solve this problem but now I'm in doubt: is $\theta$ an implicit function of $r$ or is it the other way?
$$ \ln\sqrt{x^2 + y^2} = \alpha~\arctan\frac{y}{x} $$
The implicit calculus is not so difficult than appears at first sight. We differentiate both terms :
$$\frac{x\:dx+y\:dy}{x^2+y^2}=\alpha\frac{x\:dx-y\:dy}{x^2+y^2}$$ After simplification : $$\boxed{y'=\frac{dy}{dx}=\frac{x+\alpha y}{\alpha x-y}}$$ We differentiate again :
$d(y')=\frac{\partial}{\partial x}\left(\frac{x+\alpha y}{\alpha x-y}\right)dx+\frac{\partial}{\partial y}\left(\frac{x+\alpha y}{\alpha x-y}\right)dy$
$d(y')=-\frac{(1+\alpha^2)y}{(\alpha x-y)^2}dx+\frac{(1+\alpha^2)x}{(\alpha x-y)^2}dy$
$d(y')=-\frac{(1+\alpha^2)y}{(\alpha x-y)^2}dx+\frac{(1+\alpha^2)x}{(\alpha x-y)^2}\frac{x+\alpha y}{\alpha x-y}dx$
$d(y')=(1+\alpha^2)\frac{ -y(\alpha x-y)+x(x+\alpha y)}{(\alpha x-y)^3}dx$ $$\boxed{y''=\frac{d^2y}{dx^2}=\frac{d(y')}{dx}=(1+\alpha^2)\frac{x^2+y^2}{(\alpha x-y)^3}}$$