Step 1: Name the two angles.
Let $A = \cos^{-1}\frac{3}{5}$ and $B = \tan^{-1}(-2)$. We want $\sin(A + B)$.
Step 2: Find sine and cosine of $A$.
From $\cos A = \frac{3}{5}$, draw a $3$-$4$-$5$ triangle, so $\sin A = \frac{4}{5}$ (positive, since $A$ is in $[0,\pi]$).
Step 3: Find sine and cosine of $B$.
From $\tan B = -2$, the hypotenuse is $\sqrt{1 + 4} = \sqrt 5$. Since $B$ lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$, cosine is positive: $\cos B = \frac{1}{\sqrt 5}$ and $\sin B = -\frac{2}{\sqrt 5}$.
Step 4: Apply the sine addition formula.
$\sin(A+B) = \sin A \cos B + \cos A \sin B$.
Step 5: Substitute the values.
$= \frac{4}{5} \cdot \frac{1}{\sqrt 5} + \frac{3}{5} \cdot \left(-\frac{2}{\sqrt 5}\right) = \frac{4}{5\sqrt 5} - \frac{6}{5\sqrt 5}$.
Step 6: Simplify.
$= \frac{4 - 6}{5\sqrt 5} = -\frac{2}{5\sqrt 5}$. \[ \boxed{-\dfrac{2}{5\sqrt 5}} \]