Stereographic projection is conformal in the sense of bilinear forms? This is a past exam problem from my university. However, the corresponding course sequence does not cover any Riemannian geometry, so I'm not sure how to go about this so much.

Let $S^n\subset\mathbb{R}^n$ be equipped with the round metric $g_S$, (the restriction of the Euclidean metric on $\mathbb{R}^{n+1})$. Let $H=\mathbb{R}^n\times\{0\}$ be equipped with the Euclidean metric $g_H$. Let $\Psi\colon S^n\setminus\{N\}\to H$ be stereographic projection from the North pole. Then $\Psi$ is conformal in the sense that for any $x\in S^n\setminus\{N\}$, the bilinear form $(g_S)_x$ is a multiple of the bilinear form $\Psi^\ast((g_H)_{\Psi(x)})$. 

If $\Psi\colon S^n\setminus\{N\}\to H$, the pull back sends $\Psi^\ast\colon H^2(H)\to H^2(S^n\setminus\{N\})$. I know $S^n\setminus{N}$ is diffeomorphic to $\mathbb{R}^n$, so the latter cohomology group is trivial. I think this means $(g_S)_x$ and $\Psi^\ast((g_H)_{\Psi(x)})$ are cohomologous if $(g_S)_x$ is closed? Is there a way to show that one is a multiple of the other? Thanks, I'm just curious to see what this means.

You have to do some calculations. First, find out the explicit form of $\Psi$ in terms of the euclidean coordinates of $\mathbb{R}^n$ and $\mathbb{R}^{n-1}$. You then need to calculate the derivatives of $\Psi$, which will let you write down the pull-back $\Psi^*((g_H)_{\Psi (x)})$. Indeed, as you know, the coefficients of the pulled back metric will have the form
$$
\Psi^*((g_H)_{\Psi (x)})_{ij} = \sum_{k,l} ((g_H)_{\Psi(x)})_{kl}\frac{\partial \Psi^k}{\partial x^i} \frac{\partial \Psi^l}{\partial x^j}.
$$
You just need to compute everything and compare those with the $(g_S)_{ij}$'s.
