It must be proven that the solution of the integral equation $f(x)=\int_{-\infty}^{+\infty} e^{-(x-t)^2} g(t)dt$ is $g(x)=\frac{1}{\sqrt{}\pi}\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{2^nn!} H_n(x)$
where the $H_n(x)$ are the Hermite polynomials.
It must be proven that the solution of the integral equation $f(x)=\int_{-\infty}^{+\infty} e^{-(x-t)^2} g(t)dt$ is $g(x)=\frac{1}{\sqrt{}\pi}\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{2^nn!} H_n(x)$
where the $H_n(x)$ are the Hermite polynomials.
You can start by plugging the series on the right side of the definition of $g(x)$, or rather $g(t)$, into the expression on the right side of $f(x)$. Interchange sum and integral, and you see what looks like the beginning of the Maclaurin series expansion for $f(x)$. (Your notation already assumes $f$ is infinitely differentiable.) It then suffices to show that $ x^n=\frac{1}{2^n\sqrt{\pi}}\int_{-\infty}^\infty\exp(-(x-t)^2)H_n(t)\, dt, $ as well as justifying the interchange of sum and integral.