1. The problem is to find the derivative $y'$ of the function $y = \tan^{-1}(3x)$.\n\n2. We use the formula for the derivative of the inverse tangent function: $$\frac{d}{dx} \tan^{-1}(u) = \frac{u'}{1+u^2}$$ where $u$ is a function of $x$.\n\n3. Here, $u = 3x$, so $u' = \frac{d}{dx}(3x) = 3$.\n\n4. Substitute $u$ and $u'$ into the formula: $$y' = \frac{3}{1+(3x)^2}$$\n\n5. Simplify the denominator: $$1+(3x)^2 = 1 + 9x^2$$\n\n6. Therefore, the derivative is $$y' = \frac{3}{1+9x^2}$$\n\nThis means the rate of change of $y = \tan^{-1}(3x)$ with respect to $x$ is $\frac{3}{1+9x^2}$.