Let $f$ be a non negative differentiable function (defined on $\mathbb R$) and $g(x)=\sqrt{f(x)}$. Can you characterise the points $x_0$ where $g$ is differentiable at $x_0$?
It is clear that when $f(x_0)\neq0$ then $g$ is differentiable at $x_0$ (by chain rule). So the real question is, what happens when $f(x_0)=0$? For such a point it is easy to see that if $g$ is differentiable at $x_0$ then necessarily $f'(x_0)=0$. But the converse is not true as the example $f(x)=x^2$ shows (at $x_0=0$).
So for the points $x_0$ such that $f(x_0)=0$ under what conditions on $f$ is $g$ differentiable at $x_0$? I suspect that the answer is: $g$ is differentiable at $x_0$ iff $f''(x_0)=0$, but I cannot give a proof. Do you have any ideas?
According to the Peano-Taylor theorem, if $f$ is twice differentiable at $a$ then
$$f(a+x)=f(a)+xf'(a)+\tfrac12x^2f''(a)+x^2h(x)$$
where $h$ is some function continuous at $h(0)=0$. So here's one direction of a proof:
$$f(a)=f'(a)=f''(a)=0$$ $$f(a+x)=x^2h(x)$$ $$g(a+x)=|x|\sqrt{h(x)}$$ $$g'(a)=\lim_{x\to0}\frac{g(a+x)-g(a)}{x}=\lim_{x\to0}\frac{|x|}{x}\sqrt{h(x)}=0$$
For the other direction, suppose $g(a)=g'(a)=0$:
$$f(x)=g(x)^2$$ $$f'(x)=2g(x)g'(x)$$ $$f''(a)=\lim_{x\to0}\frac{f'(a+x)-f'(a)}{x}=\lim_{x\to0}\frac{2g(a+x)g'(a+x)}{x}$$ $$=2\lim_{x\to0}\frac{g(a+x)-g(a)}{x}g'(a+x)=0$$
But this is assuming that $g'$ is defined and bounded in a neighbourhood of $a$. So here's a counter-example (let $x\geq0$ to simplify the notation, defining $g(-x)=g(x)$):
$$g(x)=x^{3/2}\sin^2x^{-1}$$ $$g(0)=g'(0)=0$$ $$f(x)=x^3\sin^4x^{-1}$$ $$f'(x)=3x^2\sin^4x^{-1}-4x\sin^3x^{-1}\cos x^{-1}$$ $$f''(0)=\lim_{x\to0}\frac{f'(x)-f'(0)}{x}=\lim_{x\to0}\Big(3x\sin^4x^{-1}-4\sin^3x^{-1}\cos x^{-1}\Big)$$
This limit doesn't exist; $4\sin^3x^{-1}\cos x^{-1}$ oscillates between $\pm3\sqrt3/4$.
Instead of referring to the ordinary second derivative, we can refer to the second Peano derivative, i.e. the coefficients of the best-approximating quadratic polynomial. The higher-order Peano derivatives of $f$ at $a$ are defined by
$$f(a+x)=\sum_{k=0}^n\tfrac{1}{k!}c_kx^k+x^nh(x),\qquad\lim_{x\to0}h(x)=0$$
It is true that the quadratic polynomial for $f$ at $a$ is $0$ if and only if the linear polynomial for $g$ at $a$ is $0$. Thus the condition you want is not $f''(a)=0$ but
$$\lim_{x\to0}\frac{f(a+x)}{x^2}=0.$$