 Differentiability of the square of a continuous function

At a point with $f(x)\ne 0$, factor the difference of squares and divide by $f(x+h)+f(x)\to 2f(x)$ to show that differentiability of $f^2$ is equivalent to that of $f$. At a zero of $f$, its square has a local minimum; if differentiable, its derivative must vanish. The difference quotient then vanishes exactly when $f(x+h{)}^2/|h|\to 0$, equivalently $f(x+h)/\sqrt{|h|}\to 0$. For example, $f(t)=|t{|}^{\alpha }$ with $1/2<\alpha <1$ is not differentiable at zero, but its square is.