Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.1 Parametric Equations - Exercises - Page 604: 68

The answer is $y=(11/7)x+311/63$.

Using Eq. (8) the slope of the tangent line is given by $\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{-27 t^2-30 t+24}{-9 t^2+6 t+6}$. So, $\frac{dy}{dx}|_{t=1/3 }=\frac{11}{7}$. At $t=1/3$, $c(1/3)=(29/9,10)$. Thus, the equation of the tangent line at $t=1/3$ is $y-10=(11/7)(x-29/9)$ So, $y=(11/7)x+311/63$.