If $y=x^3-8x+7$and $x=f\left(t\right)$ and  when  $t=0,x=3$and  $\frac{dy}{dx}=2$, then $\frac{dx}{dt}$at  $t=0$ is

$y=x^3-8x+7;x=f\left(t\right),t=0,x=3\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}\frac{dy}{dx}=2$

$x=f\left(t\right)\Rightarrow x=f\left(0\right),x=f\left(t\right)\Rightarrow \frac{dx}{dt}=f^1\left(t\right)$

$y={\left(f\left(t\right)\right)}^3-8f\left(t\right)+7$

$\frac{dy}{dx}=3.{\left(f\left(t\right)\right)}^2{f}^1\left(t\right)-8f^1\left(t\right)$

$\text{at\hspace{0.17em}\hspace{0.17em}}t=0$

$2=3{\left(f\left(0\right)\right)}^2.f^1\left(t\right)-8.f^1\left(t\right)$

$2=3{\left(3\right)}^2.f^1\left(t\right)-8.f^1\left(t\right)$

$f^{|}\left(t\right)=\frac{2}{19}$