jmc

From the condition $f^{-1}(x)=f(x),$ $f(f^{-1}(x))=f(f(x)),$ which simplifies to $f(f(x))=x.$

Note that $$\begin{array}{rl}f(f(x))& =f\left(\frac{2x+3}{kx-2}\right)\\ & =\frac{2\cdot \frac{2x+3}{kx-2}+3}{k\cdot \frac{2x+3}{kx-2}-2}\\ & =\frac{2(2x+3)+3(kx-2)}{k(2x+3)-2(kx-2)}\\ & =\frac{4x+6+3kx-6}{2kx+3k-2kx+4}\\ & =\frac{(3k+4)x}{3k+4}\\ & =x.\end{array}$$Thus, $f(f(x))=x$ for all real numbers $k,$ except when $3k+4=0,$ or $k=-4/3.$ Note that when $k=-4/3,$ $$f(x)=\frac{2x+3}{kx-2}=\frac{2x+3}{-\frac{4}{3}x-2}=\frac{3(2x+3)}{-4x-6}=\frac{3(2x+3)}{-2(2x+3)}=-\frac{3}{2},$$so $f(x)$ does not have an inverse. Hence, the answer is $k\in \boxed{(-\infty ,-\frac{4}{3})\cup (-\frac{4}{3},\infty )}.$