Suppose $(X, Y)$ is a Gaussian random vector with mean $(0, 0)$. Then $\mathbb{E}[X|Y] = \frac{\mathbb{E}[XY]}{\mathbb{E}[Y^2]}\mathbb{E}[Y]$ Suppose $(X, Y)$ is a Gaussian random vector with mean $(0, 0)$. Then $\mathbb{E}[X|Y] = \frac{\mathbb{E}[XY]}{\mathbb{E}[Y^2]}Y$ where $\mathbb{E}[X|Y]$ denotes the conditional expectation.
My guess is that this will come down to integrating the marginal and conditional densities and simplifying- I keep running into trouble when I do this however. I was also hoping to see if there was a more elegant proof- the conditional expectation looks an awful lot like a projection so I was wondering if there was anything deeper going on.
 A: Firstly some intuition. While working with conditional expectations, it's always nice to have it in form $\mathbb E[F(X,Y)|\mathcal G]$, where $X$ is independent of $\mathcal G$, $Y$ is $\mathcal G$ measurable and $F$ is borel such that $\mathbb E|F(X,Y)| < \infty$, because we have tools how to calculate those.
Having said that, let's try to make $\mathbb E[X | Y]$ looking like that above. Note that we're working with gaussian vector $(X,Y)$. Recall that if  $V=(V_1,...,V_n)$ is gaussian, then coordinates $V_1,...,V_n$ are independent if and only if covariance matrix of $V$ is diagonal (It is crucial that whole vector $V$ is gaussian, it does not work with only $V_1,...,V_n$ being gaussians).
So we can try to change our vector $(X,Y)$ into some gaussian vector (to achieve this (in general), we need a linear (or at least affine) map) $(Z,Y)$ such that $Z$ is independent of $Y$ and $X=Z+W$ where $W$ is $Y$ measurable. Since we want linear map, then $Z$ must be of form $cX+dY$, but since scalling by constant doesn't change independence (if the constant isn't zero of course) we can assume $Z=X-aY$ for some $a \in \mathbb R$. Then $W=aY$ is $Y$ measurable no matter which $a \in \mathbb R$ we choose.
To end, we need to find good $a \in \mathbb R$ meaning $X-aY,Y$ are independent. Since $(X-aY,Y)$ is gaussian, due to characterisation, it is enough to have $$ 0 = Cov(X-aY,Y) = Cov(X,Y) - aVar(Y) = \mathbb E[XY] - \mathbb E[X]\mathbb E[Y] - a\big(\mathbb E[Y^2] - (\mathbb E[Y])^2\big)  $$ and due to zero mean, we obtain $$ 0 = \mathbb E[XY] - a\mathbb E[Y^2]$$ hence $$a = \frac{\mathbb E[XY]}{\mathbb E[Y^2]}$$ (I'm assuming that $\mathbb E[Y^2] > 0$, because otherwise we just have $Y \sim \delta_0$ and such random variable is independent of $X$ by itself, hence in case $\mathbb E[Y^2] = 0$ we get $\mathbb E[X|Y] = \mathbb E[X] = 0$ (almost surely)).
Now, to end, by linearity of conditional expectation we get $$ \mathbb E[X|Y] = \mathbb E[(X-aY) + (aY) | Y] = \mathbb E[X-aY | Y] + a\mathbb E[Y|Y] $$
And since $X-aY,Y$ are independent, and $Y$ is $\sigma(Y)$ measurable, we end up with $$ \mathbb E[X|Y] = \mathbb E[X-aY] +aY = aY = \frac{\mathbb E[XY]}{\mathbb E[Y^2]}Y \quad \text{almost surely} $$
