Is it true that the covariance is 0 in this case?

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I have the joint distribution here $$f(x,y)=k\exp \left(-\frac{x^2}{100}-\left(y+\frac{3}{100}x^2-3\right)^2\right),\quad (x,y)\in \mathbb{R}^2$$ where here k is an unknown normalising constant.

I'm trying to work out the covariance between $X$ and $Y$ here. Firstly, I note that $f(x,y)$ is symmetric with respect to $x$. This means: $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}xf(x,y)\,dxdy=0$$ and therefore $E[X]=0$.

However, it also works in the below formula when I was trying to calculate $E[XY]$: $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}xyf(x,y)\,dxdy=0$$.

Therefore, this implies $Cov(X,Y)=E[XY]-E[X]E[Y]=0$ and therefore $X$ and $Y$ are independent. However, the problem is according to the factorization theorem $X$ and $Y$ are independent if and only if $f(x,y)$ can be written as $f(x)f(y)$ which is clearly not the case here.

How can I resolve this contradiction?