What function has $$\int_{-\infty}^\infty f(x) x \exp(-x^2)dx = 0$$?
The obvious answer is that $f(x)$ is an even function. Then the integral is always zero.
Another answer is $f(x) = H_k(x)$ is a Hermite polynomial with $k\neq 1$, since $H_1(x) = x$ and Hermite polynomials are orthogonal under Gaussian measure.
Are there other special functions or function classes with the above property?
Any function that has no first term in a Hermite expansion $$ f(x)=\sum_{n=0}^\infty d_n H_n(x) $$ since $$ d_n \propto \int_{-\infty}^\infty f(x)H_n(x)e^{-x^2}dx. $$
By integrating by parts, this can also be written $$ d_n \propto \int_{-\infty}^\infty \frac{\partial^n f(x)}{\partial x^n}e^{-x^2}dx $$ so the condition of no first term is equivalent to any function that satisfies $$ \int_{-\infty}^\infty \frac{\partial f(x)}{\partial x}e^{-x^2}dx=0. $$ Even functions have this property, and so do all Hermite polynomials above first order.
It is not the even-ness (symmetry) of the function that matters, we can construct a function that is not even but has this property: $$ f(x) = a\delta(x-c) + b\delta(x+d) $$ and choose $a,b,c,d$ such that $$ cae^{-a^2} - dbe^{-b^2}=0. $$
For instance, $a=1,c=1,d=2$ leads to $b\approx 1.432736294276$.
If you'd prefer a more well behaved function, replace the delta functions with smooth approximations like the Gaussian with very small variances. For instance $$ \int_{-\infty}^\infty \left(\exp\left(-\frac{(x-1)^2}{2\cdot 100000}\right)+1.43273629427692367\exp\left(-\frac{(x+2)^2}{2\cdot 100000)}\right)\right)\frac{x\exp(-x^2)}{\sqrt{2\pi\cdot 100000}}dx $$ is zero to eight decimal places.