At what point does $ y = 3ln(x)$ have maximum curvature? The formula for the curvature of a $y = f(x)$ function is given by:
$k(x) = \frac{|y'(x)|}{[1+(y''(x))^2]^\frac{3}{2}}$

Given that $y = 3ln(x),$ we know:
$y'(x) = 3/x$
$y''(x) = -3/x^2$

Plugging these in:
$k(x) = \frac{3/x}{[1+\frac{9}{x^4}]^\frac{3}{2}}$
$k(x) = \frac{3}{x[1+\frac{9}{x^4}]^\frac{3}{2}}$
$k(x) = \frac{3}{x[\frac{x^4+9}{x^4}]^\frac{3}{2}}$
$k(x) = \frac{3}{x[\frac{(x^4+9)^\frac{3}{2}}{x^6}]}$
$k(x) = \frac{3}{\frac{x}{x^6}{(x^4+9)^\frac{3}{2}}}$
$k(x) = \frac{3}{\frac{1}{x^5}{(x^4+9)^\frac{3}{2}}}$
$k(x) = \frac{3x^5}{{(x^4+9)^\frac{3}{2}}}$
From here, I need the critical points, which I get by taking the derivative of k(x):
$k'(x) = \frac{15x^4(x^4+9)^\frac{3}{2} - (3x^5)(3/2)(x^4+9)^\frac{1}{2}(4x^3)}{(x^4+9)^3}$
$k'(x) = \frac{15x^4(x^4+9)^\frac{3}{2} - 18x^8(x^4+9)^\frac{1}{2}}{(x^4+9)^3}$
$k'(x) = \frac{3x^4(x^4+9)^\frac{1}{2}[5(x^4+9)^3-6x^4]}{(x^4+9)^3}$
Beyond this, I have no idea what to do. I can solve it graphically, but how can I do it algebraically? 

The correct form of the (unsigned) curvature of a function $y(x)$ is $$\kappa(x) = \frac{|y''(x)|}{(1+(y'(x))^2)^{3/2}}.$$  For $y(x) = 3 \log x$, we then easily find $$y'(x) = \frac{3}{x}, \quad y''(x) = -\frac{3}{x^2},$$ hence $$\kappa(x) = \frac{3x}{(9+x^2)^{3/2}}, \quad x > 0.$$  It follows that $$\kappa'(x) = \frac{3(2x^2-9)}{(9+x^2)^{3/2}}$$ which has a critical point $x = 3/\sqrt{2}$.  It remains an exercise to show that this value corresponds to a local maximum of $\kappa$ and therefore the $x$-value of the point on $y = 3 \log x$ that has maximum curvature.
As an additional exercise, what is the location of the center of the osculating circle that corresponds to this maximal curvature?
As a second additional exercise, what is the locus of the centers of the osculating circles for all points on $y = 3 \log x$?  Choose a suitable parametrization.
