Here is a proof in a book I am reading. It seems fairly short, but I kind of got lost. Especially when $\lambda$ was introduced. I usually get ideas after awhile of staring at it, but I am getting nothing. I would be extremely appreciative if someone could help explain this more clearly to me or in a way I could see it better. I will keep staring at it.
Also, I don't understand why we are defining an $f'(x,y)$. What purpose does that serve?
The $\lambda$ is introduced by the definition of the center of a curve on page 54. The origin is a center when $f$ and $f'$ differ only by a non-zero scalar multiple.
Since $\sum a_{ij}x^iy^j=\sum\lambda(-1)^{i+j}a_{ij}x^iy^j$ for all $x,y$. The coefficients of every terms have to be equal, so $a_{ij}=\lambda(-1)^{i+j}a_{ij}$. This implies $1=\lambda (-1)^{i+j}$ for all $i,j$ whenever $a_{ij}\ne 0$.
Now notice that $i+j$ is also the degree of each term of the polynomial. If it could be even AND odd, you cannot have $1=\lambda (-1)^{i+j}$ for a fixed $\lambda$. So $i+j$ has to be either always even, or always odd.
For example, say in the polynomial there are terms $xy^3+xy^2+xy^5$. Here we have $i=1,j=3$ for the first term, $i=1,j=2$ for the second term, and $i=1,j=5$ for the third term. To satisfy $1=\lambda (-1)^{i+j}$ for all $i,j$, we need
$$1=\lambda (-1)^{1+3}\\ 1=\lambda (-1)^{1+2}\\ 1=\lambda (-1)^{1+5}$$
This means
$$1=\lambda\\ 1=-\lambda\\ 1=\lambda$$
which can only happen when $\lambda=0$. But by definition, $\lambda\ne 0$. Notice that if we only have the first and third term (even terms), then we can find $\lambda=1$.