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.
$$u_t - yu_x + xu_y = 0$$
$$\frac{dt}{1} = \frac{dx}{-y} = \frac{dy}{x} = \frac{du}{0} \quad \text{is OK.}$$
A first family of characteristic equations comes from $\frac{du}{0}\neq 0$ which implies
$$u=c_1$$
A second family of characteristic equations comes from $\frac{dx}{-y} = \frac{dy}{x}$ which solution is :
$$x^2+y^2=c_2$$
A third family of characteristic equations comes from $\frac{dt}{1} = \frac{dx}{-y}$ thus
$\frac{dt}{1} = \frac{dx}{-\sqrt{c_2-x^2}}$ which solution is $t=\int\frac{dx}{-\sqrt{c_2-x^2}}=-\tan^{-1}\left(\frac{x}{y}\right)+$constant.
Since $\tan^{-1}\left(\frac{x}{y}\right)=\frac{\pi}{2}-\tan^{-1}\left(\frac{y}{x}\right)$ and with a change of constant :
$$t-\tan^{-1}\left(\frac{y}{x}\right)=c_3$$
The general solution of the PDE $u_t - yu_x + xu_y = 0$ expressed on the form of implicit equation is :
$$\Phi\left(c_1\:,\: c_2\:,\: c_3 \right)=\Phi\left(u\:,\: x^2+y^2\:,\: t-\tan^{-1}\left(\frac{y}{x}\right) \right)=0$$
where $\Phi$ is an arbitrary function of three variables.
Or, on explicit form :
$$u(t,x,y)=F\left(x^2+y^2\:,\: t-\tan^{-1}\left(\frac{y}{x}\right) \right)$$
where $F$ is an arbitrary function of two variables.
Finally, $F$ has to be determined according to the boundary condition $u(0,x,y)=x+y$.
HINT : The calculus for $F$ is easier in polar coordinates.
Best Answer
$$(x-y)u_x+u_y=x$$ $$\frac{dx}{x-y}=\frac{dy}{1}=\frac{du}{x}$$ First characteristic equation from solving $\frac{dx}{x-y}=\frac{dy}{1}$ : $$(x-y-1)e^{-y}=c_1$$ $x=c_1e^y+y+1$
Second characteristic equation from solving $\frac{dy}{1}=\frac{du}{x}=\frac{du}{c_1e^y+y+1}\quad\implies\quad\frac{du}{dy}=c_1e^y+y+1$ : $$u-\frac{y^2}{2}-y-c_1e^y=c_2$$ $$u-\frac{y^2}{2}-y-((x-y-1)e^{-y})e^y=c_2$$ $$u-\frac{y^2}{2}-x+1=c_2$$ The general solution of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ with arbitrary function $F$ is : $$u-\frac{y^2}{2}-x+1=F\big((x-y-1)e^{-y} \big)$$ Explicitly the general solution of the PDE is : $$\boxed{u(x,y)=\frac{y^2}{2}+x-1+F\big((x-y-1)e^{-y} \big)}$$ Condition : $$u(x,0)=x^2=\frac{0^2}{2}+x-1+F\big((x-0-1)e^{-0} \big)$$ $$x^2-x+1=F(x-1)$$ Let $X=x-1\quad;\quad x=X+1$ $$(X+1)^2-(X+1)+1=F(X)$$ $$F(X)=X^2+X+1$$ Now the function $F$ is known. We put it into the above general solution where $X=(x-y-1)e^{-y}$ . $$u(x,y)=\frac{y^2}{2}+x-1+\big((x-y-1)e^{-y}\big)^2+\big((x-y-1)e^{-y}\big)+1$$ The particular solution of the PDE which satisfies the specified condition is : $$\boxed{u(x,y)=\frac{y^2}{2}+x+(x-y-1)^2e^{-2y}+(x-y-1)e^{-y}}$$