$\displaystyle
\begin{array}{l}
f(x) = \sqrt {1 - \sqrt {2 - \sqrt {3 - x} } } \\
\\
find \\
1)\;domain\;of\;f(x) \\
2)\;f'(x) \\
\end{array}
$
First, of course, 3- x must be non-negative in order for that square root to be a real number: $\displaystyle 3- x\ge 0$ so $\displaystyle x\le 3$. And once we have that we must have $\displaystyle 2-\sqrt{3- x}\ge 0$. That is the same as saying that $\displaystyle 2\ge \sqrt{3- x}$ or $\displaystyle 4\ge 3- x$, $\displaystyle x\ge -1$. So far then, $\displaystyle -1\le x\le 3$. Now, we need to have $\displaystyle 1- \sqrt{2- \sqrt{3- x}}\ge 0$. Can you finish that?
As for the derivative, this function is $\displaystyle y= (1- (2- (3- x)^{1/2})^{1/2})^{1/2}$. Use the "power rule" and the "chain rule".