Q:

direct differentiation in PDE

How can we check by direct differentiation that the formula $u(x, t) = \varphi(z)$, where $z$ is given implicitly by $x − z = ta(\varphi(z))$, does indeed provide a solution of the PDE $u_t + a(u)u_x = 0$?
So here's my intuition: 
find value of $z$ from $x − z = ta(\varphi(z))$, by isolating it and then substitute; I only can think of characteristics, but really can't see about the a(u)

A:

I'll just continue with nikita2's solution. To find $u_x$, different the implicit equation for $z$ with respect to $x$:
\begin{align*}
1 - z_x & = ta'(\varphi(z))\varphi'(z)z_x \\
z_x & = \frac{1}{1 + ta'(\varphi(z))\varphi'(z)}
\end{align*}
Substitute into $u_x = \varphi'(z)z_x$ and $u_t = \varphi'(z)z_t$ to get $u_t + a(u)u_x = 0$.

