= Solution
For the necessity in the <differentiability of the square of a continuous function>, suppose $g=f^2$ is <differentiable> at $x$. If $f(x)\ne0$, <continuity> makes $f(x+h)+f(x)$ nonzero for small $h$, and
$$
\frac{f(x+h)-f(x)}h
=\frac{g(x+h)-g(x)}{h[f(x+h)+f(x)]}
\longrightarrow\frac{g'(x)}{2f(x)}.
$$
Hence $f$ is <differentiable> at $x$, giving alternative (i).
If $f(x)=0$, $g(x)=0$ is a minimum because $g\geq0$. The right difference quotients of $g$ are nonnegative and the left ones are nonpositive; since their common limit exists, it must be zero. Thus $g'(x)=0$, and
$$
\frac{f(x+h)^2}{|h|}
=\operatorname{sgn}(h)\frac{g(x+h)-g(x)}h\longrightarrow0.
$$
Taking square roots of this nonnegative expression shows $f(x+h)/\sqrt{|h|}\to0$, giving alternative (ii). The two labelled parts below establish sufficiency of the respective alternatives. Together they prove the full stated equivalence, including the zero of $f$ where squaring can remove a failure of <differentiability>.
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