A curve $G$ with equation $y = \mathrm{f}(3x-2)+a$, where $a$ is a positive real constant, is transformed onto the curve with equation $y = \mathrm{f}(x)$ by a sequence of transformations.

1. Identify the scaling transformation to simplify the $x$-coefficient inside the function.

Let the original curve be $G: y = \mathrm{f}(3x-2)+a$.
To change the coefficient of $x$ from $3$ to $1$, we replace $x$ with $\frac{x}{3}$. This corresponds to a scaling parallel to the $x$-axis with a scale factor of $3$.
The equation of the intermediate curve is:
$y = \mathrm{f}\left(3\left(\frac{x}{3}\right)-2\right) + a = \mathrm{f}(x-2) + a$

2. Identify the subsequent translation to map the intermediate curve to $y = \mathrm{f}(x)$.

To transform the intermediate equation $y = \mathrm{f}(x-2) + a$ to $y = \mathrm{f}(x)$, we need to replace $x$ with $x+2$ and $y$ with $y+a$.
This corresponds to a translation by the vector $\begin{pmatrix} -2 \\ -a \end{pmatrix}$.
The equation becomes:
$y + a = \mathrm{f}((x+2)-2) + a \implies y = \mathrm{f}(x)$

1. Find the corresponding point on the curve $y = \mathrm{f}(x)$.

Since the point $(b,0)$ lies on $G$, substituting $x=b$ and $y=0$ into the equation $y = \mathrm{f}(3x-2)+a$ gives:
$0 = \mathrm{f}(3b-2) + a$
$\mathrm{f}(3b-2) = -a$
This means that on the base curve $y = \mathrm{f}(x)$, when the $x$-coordinate is $3b-2$, the $y$-coordinate is $-a$. Thus, the corresponding mapped point on $y = \mathrm{f}(x)$ is $(3b-2, -a)$.

2. Determine the coordinates $(c,d)$ on the related curve $y = \frac{1}{\mathrm{f}(x)}$.

For the related curve $y = \frac{1}{\mathrm{f}(x)}$, the point corresponding to $x = 3b-2$ will have the same $x$-coordinate, meaning:
$c = 3b-2$
The $y$-coordinate $d$ is simply the reciprocal of the $y$-value on $y = \mathrm{f}(x)$ at this $x$-coordinate:
$d = \frac{1}{\mathrm{f}(c)} = \frac{1}{\mathrm{f}(3b-2)} = \frac{1}{-a} = -\frac{1}{a}$