Put $r^2=y_1^2+y_2^2$ and write the <Riemannian metric> as a <conformal rescaling of a Riemannian metric>:
$$
g=e^{2\omega}\delta,\qquad \omega=-\frac12\log(1+r^2).
$$
The base metric is flat, so its <Ricci scalar> is zero and its <Laplacian> is $\Delta=\partial_1^2+\partial_2^2$. The formula for <scalar curvature under conformal rescaling> simplifies in dimension two to
$$
R_g=-2e^{-2\omega}\Delta\omega.
$$
Differentiate explicitly:
$$
\partial_i\omega=-\frac{y_i}{1+r^2},\qquad
\Delta\omega=-\frac2{1+r^2}+\frac{2r^2}{(1+r^2)^2}
=-\frac2{(1+r^2)^2}.
$$
Since $e^{-2\omega}=1+r^2$, the required <Ricci scalar> is
$$
\boxed{R_g(y_1,y_2)=\frac4{1+y_1^2+y_2^2}.}
$$

Solution

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

{title2=$\widetilde R=e^{-2\omega}[R-2(N-1)\Delta_g\omega-(N-1)(N-2)|\nabla\omega|_g^2]$}

In dimension $N$, the <conformal rescaling of a Riemannian metric> $\widetilde g=e^{2\omega}g$ changes the <Ricci scalar> by the displayed formula. In two dimensions the gradient-square term vanishes; for a flat base metric, $\widetilde R=-2e^{-2\omega}\Delta\omega$.

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 Incoming links: Scalar curvature under conformal rescaling