Interpretation of graph of PDE 1.Suppose we the PDE $u_{x}(x,y)=u_{y}(x,y)$. Does this simply mean we are looking for a function whos partial w.r.t $x$ and $y$ are the same att every point and then we have some additional boundary condition to get some kind of uniqueness?
2.Which in turn means that the graph of the function i.e the surface in $\mathbb{R}^{3}$ have some kind of symmetry?
3.When we have ODEs it is clear that the derivative at each point characterize the function, I have a hard time seeing the same or something similar in this case. What would be the analogy for PDE?
 A: In your case this is saying that the following directional derivative vanishes,
$$ D_v u = 0$$
where $ v = (1,-1)$. Thus the function $u$ is constant in this direction, and all such functions are therefore of the form
$$u(x,y) = f(x+y)$$
Indeed, applying the change of variables $(s,t) = (x+y,x-y)$ or equivalently $(x,y) = (\frac{s+t}2 , \frac{s-t}2)$, we see that 
$$ \partial_xu = \partial_s u + \partial_t u,\quad \partial_y u = \partial_s u - \partial_t u$$
so your PDE reduces to
$$ \partial_t u = 0$$
So your PDE (and any constant coefficient transport PDE) happens to be exactly just a collection of ODEs, one for each value of $s$. And very boring ODEs at that (constant in the direction of $t$). With regards to the first question then, You need to give me 1 initial condition for each $s$ to solve the ODE for each $s$. This determines the PDE, subject to the condition that your initial data must of course be differentiable in $s$. Also the fact that $u_x = u_y$ is more usefully visualised in the sense of the directional derivative above.
About question 3, if I understand you correctly: your PDE is special. As noted above its just a bunch of ODEs. PDEs as a whole are quite complicated and the theory is not so unified that we can make such sweeping statements, For instance there are many harmonic functions with zero boundary conditions, all of which solve $\Delta u = \sum_i \partial_i^2 u = 0$, and there are PDEs which are known to not have solutions. The heat equation has a unique smooth solution, even if you ask for very rough initial data, and it is not known how long a smooth solution to Navier-Stokes exists.
A: The fact that $u_x=u_y$ implies that $u$ is constant along all the straight lines of the form $$x+y=c.$$
Indeed, set
$$
f(t)=u(t,c-t),
$$
which describes how $f$ looks like along $x+y=c$. Then
$$
f'(t)=u_x(t,c-t)-u_y(t,c-t)=0.
$$
Hence $u$ is constant along such straight lines, and therefore
$$
u(x,y)=g(x+y).
$$
