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.
Consider fixed $c_1$ and $c_2$, then these two equations determine a curve. Now, let $c_1$ run free but remaining $c_2$ fixed, obviously we have a surface and it is a solution of the PDE. This gives us a tip about how the particular solutions emerge (we got one!). We can do better, we can move $c_1$ and simultaneously move $c_2$, we draw a surface that is solution too. But the simultaneous movement of $c_1$ and $c_2$ is what we call "function" and because this function is not still determined, we say that it is arbitrary. So $c_2=f(c_1)$ or $\psi=f(\phi)$
A particular solution determines $f$. Consider this example: we know the value of $u$ along the line $y=0$, for each $x$ we know $u$, i. e. $u$ is a perfectly known function $g$ of $x$ $u=g(x)$ along $y=0$ (you can imagine $x^2$ or $e^x$). Then each of the curves have to satisfy that requirement, so is, $\phi(x,0,g(x))=c_1$ and $\psi(x,0.g(x))=c_2$. Now, $x$ can be eliminated determining the needed functional relation between $c_1$ and $c_2$
Best Answer
The PDE $$ xu_{xx}+yu_{yy}=0 \tag{1} $$ is elliptic in the region $x<0$, $y<0$. Hence, to write it in canonical form, we need to find a change of variables $(x,y)\mapsto(\xi,\eta)$ such that$^{(*)}$ $$ w_{\xi\xi}+w_{\eta\eta}+aw_{\xi}+bw_{\eta}+cw=d, \tag{2} $$ where $w(\xi,\eta)=u(x,y)$ and $a,b,c$ and $d$ are functions of $\xi$ and $\eta$.
Let's make the simplifying assumption that $\xi=\xi(x)$ and $\eta=\eta(y)$; then, using the chain rule, we find $$ u_{xx}=\xi_x^2w_{\xi\xi}+\xi_{xx}w_{\xi}, \qquad u_{yy}=\eta_y^2w_{\eta\eta}+\eta_{yy}w_{\eta}, \tag{3} $$ so that $(1)$ becomes $$ x(\xi_x^2w_{\xi\xi}+\xi_{xx}w_{\xi})+y(\eta_y^2w_{\eta\eta}+\eta_{yy}w_{\eta})=0. \tag{4} $$ To put $(4)$ in the form $(2)$, we multiply it by $-1$ and choose $\xi$ and $\eta$ such that$^{(\dagger)}$ $\xi_x^2=-\frac{1}{x}$ and $\eta_y^2=-\frac{1}{y}$, or $(\xi,\eta)=(2(-x)^{1/2},2(-y)^{1/2})$; then $(4)$ becomes $$ w_{\xi\xi}+w_{\eta\eta}-\frac{1}{\xi}w_{\xi}-\frac{1}{\eta}w_{\eta}=0. \tag{5} $$
$^{(*)}$See, for instance, Classification of Second Order Linear PDE's and Reduction to Canonical Form, by Steve Pennell.
$^{(\dagger)}$Recall that $x<0$, $y<0$.