Example 18 Find the polynomial $f(x, y)$ of the lowest degree (but no less than 1) in $x, y$ that satisfies the following conditions:
$$
\left\{\begin{array}{l}
f(x, y) + f(y, x) = 0, \\
f(x, x + y) + f(y, x + y) = 0.
\end{array}\right.
$$
(1994 Japan Olympiad Preliminary Problem)
If $f(x, y)$ has the same degree in all its terms, i.e., for some non-negative integer $k$, we have $f(x, y)=\sum_{j=0}^{k} a_{j} x^{j} y^{k-j}$, then $f(x, y)$ is called a $k$-degree homogeneous polynomial in $x, y$. The sum of any two homogeneous polynomials of degree $i$ can be expressed as $f(x, y)=\sum_{i=0}^{n} f_{i}(x, y)$. If $f$ satisfies the two conditions given, then each $f_{i}(x, y)$ also satisfies the same conditions. Therefore, the required $f(x, y)$ is the homogeneous polynomial of the lowest degree.

From (1), we know that $f(x, y)$ is divisible by $x-y$, i.e., $f(x, y)=(x-y) h(x, y)$, where $h(x, y)$ is a homogeneous polynomial in $x, y$, and $h(x, y)=h(y, x)$, i.e., it is a symmetric polynomial.
From (2), we get $-y h(x, x+y)-x h(y, x+y)=0$,
Substituting $y-x$ for $y$ in the above equation, we get $-(y-x) h(x, y)-x h(y-x, y)=0$,
Thus, $h(x, y)$ is divisible by $x$, and by symmetry, it is also divisible by $y$, so
$$
f(x, y)=(x-y) h(x, y)=(x-y) x y g(x, y) \text {, }
$$

where $g(x, y)$ is also a homogeneous symmetric polynomial.
Applying condition (2) to the above equation, we have
$$
[x-(x+y)] x(x+y) g(z, x+y)+[y-(x+y)] y(z+y) g(y, x+y)=0 \text {, }
$$

i.e., $g(x, x+y)+g(y, x+y)=0$.
Substituting $y=-x$ into the above equation, we get $g(x, 0)+g(-x, 0)=0$.
Assume $g(x, y)$ is an $l$-degree homogeneous polynomial, i.e., $g(x, y)=\sum_{j=0}^{l} c_{j} x^{j} y^{1-j}$.

From (4), we know that $c_{l}+c_{l}(-1)^{l}=0$.
Therefore, $l$ is odd or $c_{l}=0$.
If $c_{l}=0$, then $g(x, y)$ is divisible by $y$, and by symmetry, it is also divisible by $x$, so $l \geqslant 2$. If $g(x, y)=x y$ and its constant multiples, then (3) is not satisfied, so $l \geqslant 3$.
Now consider the case where $c_{l} \neq 0$ and $l$ is odd. The first-degree symmetric polynomial $x+y$ and its constant multiples do not satisfy (3).
In summary, $l \geqslant 3$, i.e., $g(x, y)$ is at least a 3-degree homogeneous symmetric polynomial. Let
$$
g(x, y)=a\left(x^{3}+y^{3}\right)+b x y(x+y),
$$

Substituting into (3), we get $a x^{3}+a(x+y)^{3}+b x(x+y)(2 x+y)+a y^{3}+a(x+y)^{3}+b y(x+$
$$
y)(x+2 y)=0 \text {, }
$$

i.e., $a\left[\left(x^{3}+y^{3}\right)+2(x+y)^{3}\right]+b(x+y)\left[\left(2 x^{2}+x y\right)+\left(x y+2 y^{2}\right)\right]=0$,
Dividing both sides of the above equation by $x+y$, we get
$$
a\left[x^{2}-x y+y^{2}+2(x+y)^{2}\right]+b\left(2 x^{2}+2 x y+2 y^{2}\right)=0,
$$

i.e., $(3 a+2 b)\left(x^{2}+x y+y^{2}\right)=0$.
From this, we get $b=-3 a / 2$. Therefore, for example, $a=2, b=-3$ is one solution, so
$$
\begin{aligned}
f(x, y) & =(x-y) x y g(x, y) \\
& =(x-y) x y\left[\left(2 x^{3}+y^{3}\right)-3 x y(x+y)\right] \\
& =(x-y) x y(x+y)\left[2\left(x^{3}-x y+y^{2}\right)-3 x y\right] \\
& =(x-y) x y(x+y)\left(2 x^{2}-5 x y+2 y^{2}\right) \\
& =(x-y) x y(x+y)(2 x-y)(x-2 y) .
\end{aligned}
$$