I can do the parts a), b) and c) and find that in part c) that the condition in which the solution will break down is when $1+tf'(x-tu)=0$
However I am unable to part d)
I tried
$u(x,t)=f(x-tu)=(x-tu)^2=x^2+t^2u^2-2xtu$ but not sure where this should lead
This leads to a quadratic equation for $u$: $$t^2u^2 -(2xt+1)u +x^2 = 0$$ with solutions $$u(x,t) = \frac{2xt+1\pm\sqrt{(2xt+1)^2 - 4x^2t^2}}{2t^2} = \frac{2xt+1\pm\sqrt{4xt+1}}{2t^2} \tag1$$ The solution does not exist to the left of the hyperbola $x t = -1/4$. The problem of that region is that the superfast particles from $x\approx -\infty$ have enough time to get there, wreaking havoc.
To figure out the correct sign to take in (1), let $t\to 0$. If the sign is $+$ we get $\infty$, which isn't the right initial condition. So, $-$ is the only option. (It does match the initial condition; check with Taylor expansion of the square root.)