Shentao is completely correct. Therefore we have an implicit equation for $u$:

\begin{equation}

u=f(x-ut).\label{A}

\end{equation}

Remember implicit function theorem? We need $F_u\ne 0$ where $F=F(x,t,u)=u-f(x,t,u)$. Since $F(x,0,u)=u)$ we have $F_u(x,0,u)=1>0$ and (\ref{A}) is fulfilled as long $F_u(x,t,u)>0$; so solution breaks when

\begin{equation}

F_u= 1-t f(x-ut)=0.

\label{B}

\end{equation}

Let $s=x-ut$, then (\ref{A}) implies $u=f(s)$, (\ref{B}) implies $t=-1/f'(s)$, then $x=s+ut= s-f(s)/f'(s)$. So

\begin{equation}

\left\{\begin{aligned}

&x=s-f(s)/f'(s),\\

&t=f'(s)

\end{aligned}

\right.

\label{C}

\end{equation}

Figure below shows case $f(s)=-\tanh (s)$: orange lines are $u=f(s)$, $x-ut=s$ for different values of of $s$. Blue line is an

envelope of orange lines, and it shows where solution breaks. The "beak" corresponds to $s=0$ when $-f'(s)=1/cosh^2(s)$ reaches it maximum. Continuous solution exists below blue line but not above it (where orange lines overlap).