I'm trying to prove that the equation
$$u_t = p(u)\,u_x, \,\,\,\,\, t>0$$
has a solution satisfying
$$u = f(x + p(u)t)$$
where p is real.
My understanding is that one should first seek $u_t$ and $u_x$.
Having consulted my notes
$$u_t = f'(x+p(u)t) \,(p(u) + p'(u)u_t\,t)$$ $$u_x = f'(x+p(u)t)\,(1+p'(u)u_x\,t)$$
However, I do not understand how to arrive at these derivatives of $u$. If, say, we re-write the expression above as
$$u = f(g)$$
would
$$u_t = u_g \times u_t\,?$$
These are merely instances of the chain rule. For instance, $$u_t = \frac{\partial}{\partial t} f(x + p(u)t) = \frac{\partial}{\partial t}(x+p(u)t) \cdot f'(x+p(u)t) = \frac{\partial}{\partial t}(p(u)t) \cdot f'(x+p(u)t),$$ and this final derivative would be computed using the product and chain rules together: $$\frac{\partial}{\partial t}(p(u)t) = u_t \cdot p'(u) t + p(u)$$