If $y=\left|x-x^2\right|$, then $\frac{d y}{d x}$ at x = 1.

does not exist

We have,

$y=\left|x-x^2\right|= \begin{cases}x-x^2, & \text { if }-1 \leq x \leq 1 \\ x^2-x, & \text { if }|x| \geq 1\end{cases}$

Clearly, y is continuous at x = 1 but it is not differentiable at x = 1, because

(LHD at x = 1) = $\left(\frac{d}{d x}\left(x-x^2\right)\right)_{\text {at } x=1}=1-2=-1$

(RHD at x = 1) = $\left(\frac{d}{d x}\left(x^2-x\right)\right)_{\text {at } x=1}=2-1=1$

Hence, $\frac{d y}{d x}$ at x = 1 does not exist.