Answer
The maximum is $f(0)=1$ and the minimum is $f\left( 1\right) =\sqrt{2}-2 $
Work Step by Step
Given $$y=\sqrt{1+x^{2}}-2 x, \quad[0,1]$$
Since
\begin{align*}
\frac{dy}{dx}&=\frac{x}{\sqrt{1+x^2}}-2\\
&=\frac{x -2\sqrt{1+x^2} }{\sqrt{1+x^2}}
\end{align*}
Find the critical points
\begin{align*}
f'(x)&=0\\
\frac{x -2\sqrt{1+x^2} }{\sqrt{1+x^2}}&=0\\
x -2\sqrt{1+x^2}&=0\\
4-3x^2&=0
\end{align*}
Then $x=\pm2/\sqrt{3}$. Because $\pm2/\sqrt{3}\notin [0,1]$, then $f(x)$ has no critical point, since
\begin{align*}
f(0)&=1\\
f(1)&= \sqrt{2}-2
\end{align*}
Then the maximum is $f(0)=1$ and the minimum is $f\left( 1\right) =\sqrt{2}-2 $
