Use the numerator in the original PDF: it contains all $n+1$ squared differentials. The local TeX ends that numerator at $n-1$, which would not give the stated <Riemannian metric>.

For the <pullback of a Riemannian metric>, substitute the embedding $\pi(y)=(y_1,\ldots,y_n,1)$. Its differential sends a tangent vector $v$ to $(v_1,\ldots,v_n,0)$, so $dX_i=dy_i$ for $i\le n$ and $dX_{n+1}=0$. Consequently
$$
\boxed{g=\pi^*G=\frac{\sum_{i=1}^n dy_i^2}{1+\sum_{i=1}^n y_i^2},\qquad
g_{ij}(y)=\frac{\delta_{ij}}{1+|y|^2}.}
$$
The denominator is everywhere positive. For any nonzero tangent vector,
$$
g_y(v,v)=\frac{|v|^2}{1+|y|^2}>0,
$$
so this is indeed a smooth positive-definite <Riemannian metric> on $U$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 309 / 1 / b / Solution

{title2=$f^*g$}

For a smooth map $f:M\to N$, the pullback of a <Riemannian metric> is $(f^*g)_p(u,v)=g_{f(p)}(df_pu,df_pv)$. It is positive definite precisely when $df_p$ is injective at every point, that is, when $f$ is an <immersion>. In coordinates, $(f^*g)_{ij}=g_{ab}(f(y))\partial_i f^a\partial_j f^b$.

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 Incoming links: Pullback of a Riemannian metric