Modifying the problem. Rather than consider the PDE $u_{xt}+u\ u_{xx}+\frac{1}{2}u_{x}^2=0$ with initial condition $u(x,0)=u_0(x)$ as asked above, I will consider the following variant.
$$
\text{Solve }u_{xy}+u\ u_{xx} + u_x^2=0\text{ subject to }u(x,0)=f(x).\qquad(\star)
$$
There are three differences between this question and that which was asked originally.
- The coefficient of $u_x^2$ has changed from $\frac{1}{2}$ to $1$.
- The variable $t$ has been renamed to $y$.
- The initial function $u_0(x)$ has been renamed to $f(x)$.
Only (1) represents a significant modification of the problem. It makes the solution more tractable and enables it to be found using an elementary application of the method of characteristics. For these reasons, it is conceivable that this was the intended question.
Note: I will not delve into regularity of the solutions in this answer.
Reduction to a first order quasilinear PDE. Write the equation as
$$
\frac{\partial}{\partial x}\left(u_y+u\ u_x\right)=0.
$$
Thus $(\star)$ is equivalent to
$$
u_y+u\ u_x=g(y),\qquad u(x,0)=f(x),\qquad (\star\star)
$$
where $g(y)$ is an arbitrary function of $y$ (with sufficient regularity).
Method of characteristics.
Perhaps the simplest formulation of the method of characteristics is for quasilinear first order PDEs. These are PDEs of the form
$$a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u).$$
To solve this equation, one regards the solution as a surface $z=u(x,y)$ in $xyz$-space. Let $s$ parametrize the initial curve $\bigl(s,0,f(s)\bigr)$ and let $t$ be a second parameter, which can be thought of as the distance flowed along a characteristic curve emanating from $\bigl(s,0,f(s)\bigr)$.
The characteristic equations are then
$$
\frac{dx}{dt}=a(x,y,z),\quad \frac{dy}{dt}=b(x,y,z),\quad \frac{dz}{dt}=c(x,y,z).
$$
Returning to our equation $(\star\star)$, this reduces to $a(x,y,u)=u$ and $b(x,y,u)=1$ and $c(x,y,u)=g(y)$.
Thus
$$
\frac{dx}{dt}=z,\quad \frac{dy}{dt}=1,\quad \frac{dz}{dt}=g(y)
$$
with initial conditions $x(0)=s$ and $y(0)=0$ and $z(0)=f(s)$.
The solution to this system is
$$
x=s+zt,\quad y=t,\quad z=f(s)+h(t),
$$
where $h(t)$ is the antiderivative of $g(t)$ satisfying $h(0)=0$. Since $g$ was arbitrary, so is $h$ given $h(0)=0$.
The solution. Now we eliminate all occurrences of $t$ by replacing them with $y$, then eliminate $s$ by writing $s=x-zy$. Finally, replace $z$ with $u$ to obtain the implicit equation
$$
\boxed{u=f(x-uy)+h(y)},
$$
where $h(y)$ is any sufficiently regular function satisfying $h(0)=0$. This is an implicit equation for the general solution of $(\star)$.
TL;DR. Change the $\frac{1}{2}$ in the original question to $1$ to obtain a PDE solvable by the method of characteristics.
The eigenvalues look "odd" because you have mixed boundary conditions (Neumann on one side, Dirichlet on the other).
Note that the B.C. $X'(0) = 0$ forces $\color{red}{B=0}$, so the eigenfunctions have the form
$$ X_n(x) = \cos\left(\frac{(2n+1)\pi}{2L}x\right) $$
where $n = 0,1,2,\dots$ and $\lambda_n = \dfrac{\alpha\pi^2}{4L^2}(2n+1)^2$
The rest is business as usual. Write the general solution as a series and use the I.C. $u(x,0)$ to find the constants.
Edit: The solution is
$$ v(x,t) = \sum_{n=0}^\infty c_n e^{-\frac{\alpha \pi^2}{4L^2}(2n+1)^2t}\cos\left(\frac{(2n+1)\pi}{2L}x\right) $$
where
$$ c_n = \frac{\int_0^L f(x)\cos\left(\frac{(2n+1)\pi}{2L}x\right) dx}{\int_0^L \cos^2\left(\frac{(2n+1)\pi}{2L}x\right)dx } $$
and $f(x) = v(x,0)$
Edit 2: This result was derived from the orthogonality of the eigenfunctions. For $n\ne m$ you always have
$$ \int_0^L X_n(x)X_m(x)\ dx = 0 $$
To find $c_n$ such that
$$ \sum_{n=0}^\infty c_n X_n(x) = f(x) $$
Multiply both sides by $X_m(x)$ and integrate throughout
$$ \sum_{n=0}^\infty c_n\int_0^L X_n(x)X_m(x)\ dx = \int_0^L f(x)X_m(x)\ dx $$
Every term on the LHS will go to $0$ except for when $n=m$, therefore
$$ c_m \int_0^L [X_m(x)]^2\ dx = \int_0^L f(x) X_m(x)\ dx $$
Best Answer
You're almost there. Putting your solutions together, and including an arbitrary normalization constant, you have obtained the result that $$ u(x,y) = A \exp \left( - \frac{(\lambda + 2)x - y}{\lambda} \right). $$ Now plug in $y = 0$ to this form and see if you can figure out the values of $A$ and $\lambda$ needed to yield $u(x,0) = 6 e^{-3x}$.