$$
2u_{t}+t^2u_{x}=0\\
\frac{dt}{2}=\frac{dx}{t^2}=\frac{du}{0}\\
\int\frac{1}{2}t^2dt=\int1dx\\
\frac{1}{6}t^3+A=x\\
A=x-\frac{1}{6}t^3\\
$$
Because c=0, so
$$
u(t,x)=g(A)=g(x-\frac{1}{6}t^3)
$$
The initial condition given in the question: u(x,0)=f(x)
The characteristics curves ($A=x-\frac{1}{6}t^3$) will always intersect t=0 (x-axis) at a unique point, no matter what value A takes. Thus, the solution always exist.
