 Solution

Parametrize the graph by $X(u,v)=(u,v,f(u,v))$. The first fundamental form has coefficients $E=1+f_u^2$, $F=f_u{f}_v$, $G=1+f_v^2$. With unit normal $(-f_u,-f_v,1)/\sqrt{1+|\nabla f{|}^2}$, the second fundamental form has coefficients $e=f_{uu}/\sqrt{1+|\nabla f{|}^2}$, $f_{\mathrm{II}}=f_{uv}/\sqrt{1+|\nabla f{|}^2}$, and $g=f_{vv}/\sqrt{1+|\nabla f{|}^2}$. Consequently the Gaussian curvature of a graph surface is

A compact embedded surface has a highest point in the vertical direction. At that point its tangent plane is horizontal, so it is locally a graph with a local maximum of $f$. Its Hessian matrix is negative semidefinite, giving $\det D^{2}f\ge 0$ and hence $K\ge 0$. Thus the Gaussian curvature cannot be everywhere negative.

An embedded ring torus, with major radius $R$ larger than minor radius $r>0$, is a compact surface without boundary. Its standard parametrization is $((R+r\cos \theta )\cos \varphi ,(R+r\cos \theta )\sin \varphi ,r\sin \theta )$, and the Gaussian curvature of a ring torus isSince the denominator is positive, curvature is positive on the outer side and negative on the inner side, so this example changes sign.