Differentiate each of the following expressions with respect to $x$.

1. Apply the chain rule and the standard formula for the derivative of inverse cosine.

$\frac{\mathrm{d}}{\mathrm{d}x} \left( \cos^{-1}(3x^2) \right) = -\frac{1}{\sqrt{1 - (3x^2)^2}} \cdot \frac{\mathrm{d}}{\mathrm{d}x}(3x^2)$

2. Differentiate the inner quadratic term and simplify the final expression.

$= -\frac{1}{\sqrt{1 - 9x^4}} \cdot (6x)$
$= -\frac{6x}{\sqrt{1 - 9x^4}}$

1. Simplify the logarithmic expression using laws of logarithms

Let $y = \ln\left(\frac{\sqrt{2x+1}}{x^3}\right)$.
Applying the quotient and power laws of logarithms:
$y = \ln(2x+1)^{\frac{1}{2}} - \ln(x^3)$
$y = \frac{1}{2}\ln(2x+1) - 3\ln x$

2. Differentiate term by term with respect to $x$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2} \left( \frac{2}{2x+1} \right) - 3 \left( \frac{1}{x} \right)$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2x+1} - \frac{3}{x}$

3. Combine the result into a single simplified fraction

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x - 3(2x+1)}{x(2x+1)}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x - 6x - 3}{x(2x+1)}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{5x + 3}{x(2x+1)}$