Statement I: Let S be the set of points where the function $f\left(x\right)=\left|2-\left|x-3\right|\right|,\forall x\in R$ is not differentiable then $\sum _{x\in S}^{ }f\left(f\left(x\right)\right)$ ,is equal to 3
Statement II:   . If $\sqrt{\frac{y}{x}}+4\sqrt{\frac{x}{y}}=4$ then $\frac{dy}{dx}=4$
 

Statement : I

$\mathrm{f}\left(\mathrm{x}\right)=|2-|\mathrm{x}-3\parallel ,\mathrm{x}\in \mathrm{R} \text{is not differentiable if }|\mathrm{x}-3|=0 \text{and}$

$2-|\text{x}-3|=0\Rightarrow \text{x}=3\text{ and }|\text{x}-3|=2\Rightarrow \text{x}-3=2,-2$

$\Rightarrow \text{x}=5,1\therefore \text{s}=\{1,3,5\}\text{f}\left(1\right)=|2-2|=0\Rightarrow \text{f}\left(\text{f}0\right))=\text{f}(0)=1$

$,\text{f}\left(3\right)=|2-0|=2\Rightarrow \text{f}\left(\text{f}\right(2\left)\right)=\text{f}\left(2\right)=1\text{, and f}\left(5\right)=|2-2|=0\Rightarrow \text{f}\left(\text{f}\right(5\left)\right)=\text{f}\left(0\right)=1$

$\Rightarrow \sum _{x\in 5}\left(f\right(x\left)\right)=3$

Statement : II

$\begin{array}{c}\sqrt{\frac{y}{x}}+4\sqrt{\frac{x}{y}}=4\\ \text{ Let }\sqrt{\frac{y}{x}}=t\\ \Rightarrow t+\frac{4}{t}=4\Rightarrow t^2-4t+4=0\\ \therefore (t-2{)}^2=0\Rightarrow t=2\\ \sqrt{\frac{y}{x}}=2\Rightarrow \frac{y}{x}=4\\ \therefore y=4x\Rightarrow \frac{dy}{dx}=4\end{array}$

$f\left(x\right)=|2-|x-3\parallel ,x\in R$

is not differentiable at 1,3,5