Gaussian Curvature of Saddle

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I recently got to grips with simple 2D curvature, and I'm aware that formulae exist for Gauss curvatures of surfaces defined in terms of some $z=f(x,y)$, but I'm finding little progress making sense of these. Is there a nice way to get at the curvature of such defined surfaces?

In particular I'd like to find the curvatures of the saddles defined by $z=xy$ and $z=x^y\;(0^0:=1)$ (intuitively, the latter "feels" more negative than the former, but by how much?).

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Here is cooking recipe for computing Gaussian and Mean curvatures.
1. First you want to parametrise the surface, naive $\phi(u,v)=(u,v,uv)$ will do, or equivalently $\phi(u,v)=(u,v,u^2-v^2)$ for same surface. (Similiarly $\psi(u,v)=(u,v,u^v)$ for surface defined by $z=x^y$ with saddle point at $\psi(1,0)$.)
2. Compute first partial derivatives: $\partial_u\phi, \partial_v\phi$.
3. While you are at it compute second partial derivatives: $\partial^2_{uu}\phi, \partial^2_{uv}\phi,\partial^2_{vv}\phi$.
4. Compute coefficients of first fundamental form by
$g_{11}=\langle\partial_u\phi,\partial_u\phi \rangle$,
$g_{12}=g_{21}=\langle\partial_u\phi,\partial_v\phi \rangle$
$g_{22}=\langle\partial_v\phi,\partial_v\phi \rangle$
,where $\langle ,\rangle$ denotes standard scalar product of $\mathbb R^3$
5. Now you need a normal unit field, which you can construct from 2.: $$N=\frac{\partial_u\phi \times \partial_v\phi}{\Vert \partial_u\phi \times \partial_v\phi \Vert} $$ where $\times$ denotes cross product.
6. Compute coefficients of second fundamental form by
$h_{11}=\langle\partial^2_{uu}\phi,N \rangle$,
$h_{12}=h_{21}=\langle\partial^2_{uv}\phi,N \rangle$
$h_{22}=\langle\partial^2_{vv}\phi,N \rangle$
7. Now you are ready to compute Gaussian curvature: $$ K=\frac{h_{11}\cdot h_{22}-h_{12}^2}{g_{11}\cdot g_{22}-g_{12}^2}$$ and mean Curvature $$H=\frac{1}{2}\frac{h_{11}\cdot g_{22}-2\cdot h_{12}\cdot g_{12}+h_{22}\cdot g_{11}}{g_{11}\cdot g_{22}-g_{12}^2} $$

If done right, you should get $K(\phi(0,0))=-4$ for $z=xy$ surface.