Does strict convexity imply differentiability?

$$f(x)=x^2+|x|$$ is strictly convex because of the $x^2$ term but not differentiable at $0$ because of the $|x|$ term

Choose a strictly convex function $u$ and some sequences $(x_n)$ and $(a_n)$ such that every $a_n$ is positive and the series $\sum\limits_na_n(1+|x_n|)$ converges. Then the formula $$ v(x)=\sum\limits_na_n|x-x_n| $$ defines a proper function $v$ such that $u+v$ is strictly convex, and not differentiable at any $x_n$.

The countable set of points $X=\{x_n\}$ may be dense. The function $v$ is differentiable at every $x$ not in $X$, with $$ v'(x)=\sum\limits_na_n\,\mathrm{sgn}(x-x_n). $$

The function $$f(x)=\max(e^x,e^{-x})$$ is strictly convex but not differentiable at $0$.