The equation of a curve is given by $y = \ln \sqrt{\dfrac{3x^3}{2x-1}}$, where $x > \tfrac{1}{2}$.

1. Simplify the logarithmic expression before differentiating.

$y=\ln\sqrt{\frac{3x^3}{2x-1}}$
$y=\frac12\ln\left(\frac{3x^3}{2x-1}\right)$
$y=\frac12\left(\ln 3+3\ln x-\ln(2x-1)\right)$

2. Differentiate $y$ with respect to $x$.

$\frac{dy}{dx}=\frac12\left(0+\frac{3}{x}-\frac{2}{2x-1}\right)$
$\frac{dy}{dx}=\frac12\left(\frac{3}{x}-\frac{2}{2x-1}\right)$

3. Use the condition for a stationary point to find $x$.

At a stationary point, $\frac{dy}{dx}=0.$

So
$\frac12\left(\frac{3}{x}-\frac{2}{2x-1}\right)=0$
$\frac{3}{x}=\frac{2}{2x-1}$
$3(2x-1)=2x$
$6x-3=2x$
$4x=3$
$x=\frac34$

Since $\frac34>\frac12$, this value is in the given domain.

4. Substitute the $x$-coordinate into the curve equation to find $y$.

When $x=\frac34$,
$y=\ln\sqrt{\frac{3\left(\frac34\right)^3}{2\left(\frac34\right)-1}}$
$=\ln\sqrt{\frac{3\cdot \frac{27}{64}}{\frac32-1}}$
$=\ln\sqrt{\frac{\frac{81}{64}}{\frac12}}$
$=\ln\sqrt{\frac{81}{32}}$

1. Use the first derivative from part (i) and differentiate again.

From part (i),
$\frac{dy}{dx}=\frac{3}{2x}-\frac{1}{2x-1}.$
Hence
$\frac{d^2y}{dx^2}=-\frac{3}{2x^2}+\frac{2}{(2x-1)^2}.$

2. Evaluate the second derivative at the stationary point.

At the stationary point, $x=\dfrac{3}{4}$.
$\frac{d^2y}{dx^2}=-\frac{3}{2\left(\frac{3}{4}\right)^2}+\frac{2}{\left(2\left(\frac{3}{4}\right)-1\right)^2}$
$=-\frac{3}{2\cdot\frac{9}{16}}+\frac{2}{\left(\frac{1}{2}\right)^2}$
$=-\frac{8}{3}+8=\frac{16}{3}.$

3. Conclude the nature using the second derivative test.

Since
$\frac{d^2y}{dx^2}=\frac{16}{3}>0,$
the curve is concave upwards at the stationary point.