$\textbf{Set-up.}$ Let $g$ be some $L^2$ function such that for $i+j = 2$ $$ \frac{x^iy^j}{g(x,y)}\in L^2(B), $$ where $B$ is some open ball centered at $(0,0)$ of some radius $\delta$. Specifically, $$ \frac{x^2}{g(x,y)},\frac{xy}{g(x,y)}, \frac{y^2}{g(x,y)}\in L^2(B). $$ Moreover, $$ \frac{1}{g(x,y)},\frac{x}{g(x,y)},\frac{y}{g(x,y)}\not\in L^2(B) $$ for any ball $B$.
$\textbf{Claim.}$ If $$ \frac{c_1 x + c_2 y}{g(x,y)}\in L^2(B), $$ then $c_1 = c_2 = 0$.
$\textbf{Ideas.}$ If $g$ is the Euclidean-distance function, this result follows as one is able to convert to polar coordinates and things are nice. In this more abstract setting, it is much more unclear. My intuition is saying that this claim should be true as any linear combination of these functions is not contributing an addition "zero" that could control the blowup of $\frac{1}{g(x,y)}$. In general, there are linear combinations of non-integrable functions that are integrable. For example, $$ \frac{x}{\log(x)},\frac{-1}{\log(x)}\not\in L^1[0,1], $$ but $$ \frac{x-1}{\log(x)}\in L^{\infty}[0,1]. $$ However, in this case the top function has a "new" zero that can control the blowup of $\frac{1}{\log(x)}$ at $x = 1$. In my case, this cannot happen so I am lead to believe I'm correct, but I cannot prove said result. Any advice/direction would be appreciated. Or a counterexample would be nice too if I'm wrong.
A counter example is as follows. Let $$g(x,y) = (x^2+y^2)^{\frac{1+\frac{2xy}{x^2+y^2}}{2}}.$$ Note that if we convert to polar coordinates, we have that $$\widetilde{g}(r,\theta) = r^{1+\sin(2\theta)}.$$ One can show that for $i+j = 2$ $$ \frac{x^iy^j}{g(x,y)}\in L^2(B), $$ whereas for $i+j = 1$, we have that $$ \frac{x^iy^j}{g(x,y)}\not\in L^2(B). $$ Lastly, one can show (through some slightly tedious computations) that $$ \frac{x-y}{g(x,y)}\in L^2(B). $$