A smooth function $f(x)$ has a unique local and global minimum. What happens to its location as $f(x)$ varies smoothly in time?

For fixed $t$, $f(x,t) = (x^3−t)^2$ has its unique local and global minimum at $x = t^{1/3}$. This is not differentiable at $t=0$.

I am not sure about the differentiability of $\chi$, but unless I missed something it is really easy to show that $\chi$ is continuous if $f$ is.

Indeed, consider $t_0\in [0,1]$ and let $k_0=\chi(t_0)$. Let $\varepsilon>0$, and $g(t)= \min(f(k_0-\varepsilon,t)-f(k_0,t),f(k_0+\varepsilon,t)-f(k_0,t))$. Then $g$ is continuous since $f$ is, and $g(t_0)>0$, so there must be a $\eta>0$ such that $g(t)>0$ for any $t\in[t_0-\eta,t_0+\eta]\cap [0,1]$. Consider now, for such $t$, the restriction of $F_t(x)=f(x,t)$ to $[k_0-\varepsilon,k_0+\varepsilon]$ ; by compactness, there is a $z\in [k_0-\varepsilon,k_0+\varepsilon]$ where $F_t$ reaches its minimum. Since $z\leq F_t(k_0)$, $z$ must be strictly inside $[k_0-\varepsilon,k_0+\varepsilon]$. Because of the hypotheses on $f$, this forces $z$ to coincide with $\chi(t)$. This shows that $|\chi(t)-\chi(t_0)|<\varepsilon$ whenever $|t-t_0|<\eta$, and this is the classic definition of continuity.