The function $f(x) = {e^{|x|}}$ is

Let $u(x) = |x|$ and $v(x) = {e^x}$

therefore,$f(x) = {\mathop{\rm vou}\nolimits} (x) = v[u(x)]$
$= v|x| = {e^{|x|}}$

Since, $u(x)$ and $v(x)$ are both continuous functions.

So, $f(x)$ is also continuous function but $u(x) = |x|$ is not differentiable at $x = 0,$ whereas $v(x) = {e^x}$ is differentiable at everywhere.

Hence, $f(x)$ is continuous everywhere but not differentiable at $x = 0$.