Let a function $y=y\left(x\right)$  be defined parametrically by $x=2t-\left|t\right|$ , $y=t^2+t\left|t\right|$  then

$ifx>0$

$y=t^2+t^2\left(t>0\right)$

$y=2t^2$

$\frac{dy}{dt}=4t$

$t>0\Rightarrow x=2t-t=t\left(t>0\right)$

$\frac{dx}{dt}=1$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{4t}{1}=4t=4x,x>0$

$y^I\left(x\right)=4x,whenx>0$

$x<0\Rightarrow x=2t+t=3t$

$\frac{dx}{dt}=3$

$t<0\Rightarrow y=t^2+t\left(-t\right)=0$

$\frac{dy}{dt}=0$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{0}{3}=0itx<0$