Breakdown of solution to inviscid Burgers equation 
Let $u = f(x-ut)$ where $f$ is differentiable. Show that $u$ (amost always) satisfies $u_t + uu_x = 0$. What circumstances is it not necessarily satisfied?

This is a question in a tutorial sheet I have been given and I am slightly stuck with the second part. To show that $u$ satisfies the equation I have differentiated it to get:
$u_t = -f'(x-ut)u$
$u_x = f'(x-ut)$
Then I have substituted these results into the original equation. The part I am unsure of is where it is not satisfied. If someone could push me in the right direction it would be much appreciated.

We have 
\[
  u_t = f'(x-ut)(x-ut)_t = -f'(x-ut)(u_t t + u) 
     \iff \bigl(1 + tf'(x-ut)\bigr)u_t = -uf'(x-ut) 
\]
and
\[
   u_x = f'(x-ut)(x-ut)_x = f'(x-ut)(1 - u_xt)
  \iff \bigl(1 + tf'(x-ut)\bigr)u_x = f'(x-ut) 
\]
Which gives that 
\[
  \bigl(1 + tf'(x-ut)\bigr)(u_t +uu_x) = 0
\]
so at each point either $1 + tf'(x-ut) = 0$ or $u_t + uu_x = 0$.
