Along the characteristic curves:

\begin{gather*}

\frac{dt}{1} = \frac{dx}{t^2+1} \\

\int{t^2+1 \, dt} = \int{dx}\\

\frac{t^3}{3} + t + C = x

\end{gather*}

Also $f(x,t,u) = 0$ implies $u$ is constant along the characteristic curves,

therefore:

\begin{equation*}

u = \phi(x-\frac{t^3}{3}-t)

\end{equation*}