a) Step 1: Differentiate $\ln\left(\frac{3x^3}{\sin x}\right)$ with respect to $x$. First, simplify the expression using logarithm properties: $$ \ln\left(\frac{3x^3}{\sin x}\right) = \ln(3x^3) - \ln(\sin x) $$ $$ = \ln(3) + \ln(x^3) - \ln(\sin x) $$ $$ = \ln(3) + 3\ln(x) - \ln(\sin x) $$ Now, differentiate each term with respect to $x$: $$ \frac{d}{dx}\left(\ln(3) + 3\ln(x) - \ln(\sin x)\right) $$ $$ = \frac{d}{dx}(\ln(3)) + \frac{d}{dx}(3\ln(x)) - \frac{d}{dx}(\ln(\sin x)) $$ $$ = 0 + 3\left(\frac{1}{x}\right) - \left(\frac{1}{\sin x} \cdot \cos x\right) $$ $$ = \frac{3}{x} - \frac{\cos x}{\sin x} $$ $$ = \frac{3}{x} - \cot x $$ The derivative is $\boxed{\frac{3}{x} - \cot x}$. b) Step 1: Find $\frac{dy}{dx}$ for the parametric curve. The curve is defined by $x = t^3$ and $y = t - \frac{1}{4}t^3$. First, find $\frac{dx}{dt}$ and $\frac{dy}{dt}$: $$ \frac{dx}{dt} = \frac{d}{dt}(t^3) = 3t^2 $$ $$ \frac{dy}{dt} = \frac{d}{dt}\left(t - \frac{1}{4}t^3\right) = 1 - \frac{1}{4}(3t^2) = 1 - \frac{3}{4}t^2 $$ Now, find $\frac{dy}{dx}$ using the chain rule: $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1 - \frac{3}{4}t^2}{3t^2} $$ Step 2: Find the gradient of the tangent at $t = -1$. Substitute $t = -1$ into the expression for $\frac{dy}{dx}$: $$ m = \frac{1 - \frac{3}{4}(-1)^2}{3(-1)^2} = \frac{1 - \frac{3}{4}(1)}{3(1)} = \frac{1 - \frac{3}{4}}{3} = \frac{\frac{1}{4}}{3} = \frac{1}{12} $$ Step 3: Find the coordinates $(x, y)$ of the point on the curve at $t = -1$. Substitute $t = -1$ into the parametric equations for $x$ and $y$: $$ x = (-1)^3 = -1 $$ $$ y = (-1) - \frac{1}{4}(-1)^3 = -1 - \frac{1}{4}(-1) = -1 + \frac{1}{4} = -\frac{4}{4} + \frac{1}{4} = -\frac{3}{4} $$ The point of tangency is $\left(-1, -\frac{3}{4}\right)$. Step 4: Write the equation of the tangent line. Using the point-slope form $y - y_1 = m(x - x_1)$: $$ y - \left(-\frac{3}{4}\right) = \frac{1}{12}(x - (-1)) $$ $$ y + \frac{3}{4} = \frac{1}{12}(x + 1) $$ To eliminate fractions, multiply the entire equation by 12: $$ 12\left(y + \frac{3}{4}\right) = 12\left(\frac{1}{12}(x + 1)\right) $$ $$ 12y + 9 = x + 1 $$ Rearrange the equation into the form $Ax + By + C = 0$: $$ x - 12y + 1 - 9 = 0 $$ $$ x - 12y - 8 = 0 $$ The equation of the tangent to the curve is $\boxed{x - 12y - 8 = 0}$.