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?