P(x,y)

[Navid]

Find all Symmetric Polynomials $P(x,y)$ such that $ P(x,y)=P(x,x-y)$.

[ИСН]

Well, "symmetric" means that $(x,y)\mapsto(y,x)$. Your condition means that $(x,y)\mapsto(x,x-y)$. Now let's take all combinations of these two (they happen to generate a dihedral group of order 12) and get to the result.
Let $Q(x,y)$ be any polynomial; then

$$\begin{align}Q(x, y) + Q(x, x - y) + Q(y, x) + Q(x - y, x) + Q(y, -x + y) + Q(x - y, -y) +\\+ Q(-x + y, y) + Q(-y, x - y) + Q(-x + y, -x) + Q(-x, -x + y) + Q(-y, -x) + Q(-x, -y)\end{align}$$

is your $P(x,y)$. The obvious implications as to the degree of $P$ I leave to yourself.

$x^2-xy+y^2$ seems to be the "simplest" non-trivial example in some sense.