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.

Following @martini's comment, and rearranging, you should find
$$ (1+tf')u_t=-uf',$$
omitting the argument of $f'$ for convenience. Likewise, 
$$ (1+tf')u_x=f'.$$
These lead to 
$$(1+tf')(u_t+uu_x)=0,$$ from which the equation $u_t+uu_x=0$ follows, provided the coefficient $1+tf'(x-ut)$ is non-zero. This is the source of the 'almost always' aspect of the result.
