= Solution
For
$$
V=4t^2\partial_t+4tx\partial_x-(x^2+2t)\partial_u,
$$
the flow equations are
$$
\frac{dT}{d\epsilon}=4T^2,
\qquad
\frac{dX}{d\epsilon}=4TX,
\qquad
\frac{dU}{d\epsilon}=-(X^2+2T),
$$
with $(T,X,U)=(t,x,u)$ at $\epsilon=0$. Solving the first two gives, with $d=1-4\epsilon t$,
$$
T=\frac td,
\qquad X=\frac xd.
$$
Integrating the last equation then gives the one-parameter group
$$
\boxed{
T=\frac{t}{1-4\epsilon t},
\qquad
X=\frac{x}{1-4\epsilon t},
\qquad
U=u-\frac{\epsilon x^2}{1-4\epsilon t}
+\frac12\log(1-4\epsilon t),}
$$
defined locally where $1-4\epsilon t>0$.
To verify the symmetry infinitesimally, use
$$
\xi^t=4t^2,
\qquad \xi^x=4tx,
\qquad \eta=-(x^2+2t).
$$
The needed coefficients of the <second prolongation of a Lie point symmetry> are
$$
\eta^t=-2-8tu_t-4xu_x,
\qquad
\eta^x=-2x-4tu_x,
$$
and
$$
\eta^{xx}=-2-8tu_{xx}.
$$
For the <potential Burgers equation> written as
$$
F=u_t-u_{xx}-u_x^2=0,
$$
we obtain
$$
\begin{aligned}
\operatorname{pr}^{(2)}V(F)
&=\eta^t-\eta^{xx}-2u_x\eta^x\\
&=-8t(u_t-u_{xx}-u_x^2)\\
&=-8tF.
\end{aligned}
$$
\b[Thus the prolonged generator is tangent to the solution manifold $F=0$, and the finite transformations above form the <projective Lie symmetry of the potential Burgers equation>.]
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