For what values of $x$is the rate of increase of $x^3-5x^2+5x+8$ is twice the rate of increase of $x$ ?

Let $y=x^3-5x^2+5x+8$. Then,

$\frac{dy}{dt}=\left(3x^2-10x+5\right)\frac{dx}{dt}$

When $\frac{dy}{dt}=2\frac{dx}{dt}$ , We have

$\begin{array}{l} \left(3x^2-10x+5\right)\frac{dx}{dt}=2\frac{dx}{dt}\\ \Rightarrow 3x^2-10x+3=0\\ \Rightarrow (3x-1)(x-3)=0\\ \Rightarrow x=3,\frac{1}{3}\end{array}$