1. **State the problem:** Find the derivative $\frac{d}{dx} \csc^{-1}(2x)$ at $x=\frac{3}{5}$.\n\n2. **Recall the formula:** The derivative of $y = \csc^{-1}(u)$ with respect to $x$ is given by $$\frac{dy}{dx} = -\frac{1}{|u| \sqrt{u^2 - 1}} \cdot \frac{du}{dx}.$$\nHere, $u = 2x$, so $\frac{du}{dx} = 2$.\n\n3. **Apply the formula:** Substitute $u=2x$ and $\frac{du}{dx}=2$ into the formula: $$\frac{d}{dx} \csc^{-1}(2x) = -\frac{1}{|2x| \sqrt{(2x)^2 - 1}} \cdot 2 = -\frac{2}{|2x| \sqrt{4x^2 - 1}}.$$\n\n4. **Simplify the expression:** Note that $|2x| = 2|x|$, so $$\frac{d}{dx} \csc^{-1}(2x) = -\frac{2}{2|x| \sqrt{4x^2 - 1}} = -\frac{\cancel{2}}{\cancel{2} |x| \sqrt{4x^2 - 1}} = -\frac{1}{|x| \sqrt{4x^2 - 1}}.$$\n\n5. **Evaluate at $x=\frac{3}{5}$:** Calculate inside the square root: $$4x^2 - 1 = 4 \left(\frac{3}{5}\right)^2 - 1 = 4 \cdot \frac{9}{25} - 1 = \frac{36}{25} - 1 = \frac{36}{25} - \frac{25}{25} = \frac{11}{25}.$$\n\n6. **Calculate the derivative value:** $$\frac{d}{dx} \csc^{-1}(2x) \bigg|_{x=\frac{3}{5}} = -\frac{1}{\left| \frac{3}{5} \right| \sqrt{\frac{11}{25}}} = -\frac{1}{\frac{3}{5} \cdot \frac{\sqrt{11}}{5}} = -\frac{1}{\frac{3 \sqrt{11}}{25}} = -\frac{25}{3 \sqrt{11}}.$$\n\n7. **Rationalize the denominator:** $$-\frac{25}{3 \sqrt{11}} \cdot \frac{\sqrt{11}}{\sqrt{11}} = -\frac{25 \sqrt{11}}{3 \cdot 11} = -\frac{25 \sqrt{11}}{33}.$$\n\n**Final answer:** $$\boxed{-\frac{25 \sqrt{11}}{33}}.$$