= Solution
An $n$-node <Gaussian quadrature> formula chooses nodes and weights for an integral with a positive weight so that every <polynomial> of degree at most $2n-1$ is integrated exactly. Its nodes are the zeros of the degree-$n$ <orthogonal polynomial> for that weight.
For the even weight $w(x)=1-x^2$, the monic orthogonal quadratic has the form $p_2(x)=x^2-d$. Orthogonality to $x$ follows from oddness; orthogonality to one gives
$$
d=\frac{\int_{-1}^1x^2(1-x^2)dx}{\int_{-1}^1(1-x^2)dx}=\frac{4/15}{4/3}=\frac15.
$$
The nodes are $\pm1/\sqrt5$. The equal weights sum to $4/3$, hence each is $2/3$. The <two-node Gaussian quadrature with quadratic weight> is therefore
$$
\boxed{\int_{-1}^1(1-x^2)f(x)dx\simeq\frac23\left[f(-1/\sqrt5)+f(1/\sqrt5)\right].}
$$
To verify cubic exactness, divide any cubic as $P=p_2Q+R$ with $\deg Q,\deg R\leq1$. Orthogonality removes $p_2Q$ from the integral and its nodal values vanish; linear interpolation integrates $R$ exactly. This also proves the Gaussian property rather than merely matching two moments.
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