Conditional expectation $E(X\mid XY)$ when $(X,Y)$ is standard normal I have to calculate 
$\mathbb{E}(X|X*Y)$ with X,Y being independent and standard normal distributed. I got at tip in this post (Conditional expectation on components of gaussian vector), that I should use the definition and Bayes Thm to solve the problem. I played around a bit, but I just don't get it :(

May anyone give me another hint?
As every conditional expectation, $E[X\mid XY]=w(XY)$ for some measurable function $w$. Recall that:


  Conditional expectations depend only on the joint distribution of the random variables considered, in the sense that, if $E[X\mid XY]=w(XY)$, then $E[X'\mid X'Y']=w(X'Y')$ for every $(X',Y')$ distributed like $(X,Y)$. 


Choosing $(X',Y')=(-X,-Y)$ above, one gets $X'Y'=XY$ hence $$w(XY)=E[-X\mid XY]=-E[X\mid XY]=-w(XY).$$ Thus, $$E[X\mid XY]=0.
$$
One sees that $E[X\mid XY]=0$ for every centered gaussian vector $(X,Y)$, neither necessarily independent nor standard.

Still more generally:


  Let $\Xi$ denote any centered gaussian vector, $u$ an odd measurable function such that $u(\Xi)$ is integrable and $v$ an even measurable function. Then, $$E[u(\Xi)\mid v(\Xi)]=0.$$
you need to compute the integral : 
$$ \int_{-\infty}^{\infty} (x f_{x|xy}) dx $$
where $f_{x|xy}$ is the conditional pdf. To find this pdf, $$f_{x|xy} = \frac{f_{x,xy}}{f_{xy}}$$ 
Try to do a transformation with $U=XY$ and $V=.....$ that can help you to compute the expectation