= Solution
Set $D=1+u^2+v^2$. The two tangent vectors to the <Enneper surface> are
$$
\sigma_u=(1-u^2+v^2,2uv,2u),\qquad\sigma_v=(2uv,1-v^2+u^2,-2v).
$$
Their <cross product> is $D(-2u,2v,1-u^2-v^2)$. Since $4u^2+4v^2+(1-u^2-v^2)^2=D^2$, its length is $D^2>0$, proving regularity and giving the <unit normal>
$$
\boxed{\mathbf n=\frac1D(-2u,2v,1-u^2-v^2).}
$$
For the <first fundamental form>, direct dot products give $E=G=D^2$ and $F=0$. The second derivatives are $\sigma_{uu}=(-2u,2v,2)$, $\sigma_{uv}=(2v,2u,0)$, and $\sigma_{vv}=(2u,-2v,-2)$. Taking their dot products with $\mathbf n$ gives the <second fundamental form> coefficients $e=2$, $f=0$, $g=-2$. Therefore the <Gaussian curvature> is
$$
\boxed{K=\frac{eg-f^2}{EG-F^2}=-\frac4{(1+u^2+v^2)^4}.}
$$
Here curvature means <Gaussian curvature>; the opposite signs of the second-form diagonal entries also give zero <mean curvature>, as expected for this <minimal surface>.
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