Finding the moment generating function of the product of two standard normal distributions

The MGF is, by definition, $M(t) = E[e^{tUV}]$. Try integrating $$ \dfrac{1}{2 \pi} \int_{-\infty}^\infty \int_{-\infty}^\infty e^{tuv} e^{-u^2/2} e^{-v^2/2}\ du\ dv$$

by completing the square.

There's a nice way to do this avoiding any tricky integration that I learned from here: observe $XY = (1/4)(X+Y)^2 - (1/4)(X-Y)^2$ and that $X+Y$ and $X-Y$ are independent, so $XY$ is the sum of scalar multiples of $\chi^2$ random variables with one degree of freedom. Using the fact that the MGF of a $\chi^2_1$ random variable is $(1-2t)^{-1/2}$ you can now get the MGF you want.