help

To find the inverse of f(x), set f(x) = y. Then solve for x in terms of y. 

By doing this I got $x=\frac{2y+3}{ky-2}$, so $f^{-1}(x)=\frac{2x+3}{kx-2}$.

This means that $k$ can be any real number for $f(x) = f^{-1}(x)$ (as long as $x\ne\frac{2}{k}$).

We can write

y = [ 2x + 3 ] / [ kx - 2]        isolate x

y [ kx - 2 ] =  2x + 3

(ky)x - 2y = 2x + 3

(ky)x - 2x =  2y + 3

(ky - 2) x =  2y + 3    

x =  (2y + 3) / ( ky - 2)    "swap" x and y

y = (2x + 3) / (kx - 2)  =  the inverse

The inverse is the same as f(x)

So.....k can take on any real value

help
Let

$\large{f(x) = \dfrac{2x + 3}{kx - 2}}$.

Find all real numbers

$\mathbf{k}$

so that

$\large{f^{-1}(x) = f(x)}$.

discontinuity:

$\begin{array}{|rcll|} \hline 2x+3 &=& 0 \\ 2x &=& -3 \\ x &=& -\dfrac{3}{2} \\ \hline kx - 2 &\ne& 0 \\ kx &\ne& 2 \\ k &\ne& \dfrac{2}{x} \quad | \quad x = -\dfrac{3}{2} \\\\ k &\ne& \dfrac{2}{-\dfrac{3}{2}} \\\\ \mathbf{k} & \mathbf{\ne} & \mathbf{-\dfrac{4}{3}} \\ \hline \end{array}$

At $k = -\dfrac{4}{3} \Rightarrow x = -\dfrac{3}{2} \Rightarrow y = \dfrac{0}{0}$