Understanding mean curvature and Weingarten maps

Question

The parametrized surface: $$\textbf{R}\supset{U}\ni(x,y)\longmapsto (x,y,f(x,y))\in\textbf{R}^3$$ defined by the graph of a smooth function $z=f(x,y)$ I know that the Weingarten map:$$L:=II\circ I^{-1}$$ equals the Guassian curvature. Now I'm interested in calculating that the trace of L (which I believe is the mean curvature of this surface) and using that to prove that the surface has mean curvature zero if and only if $f$ satisfies the minimal surface equation, namely $$H=\frac{(1+h_y^2)h_{xx}-2h_xh_yh_{xy}+(1+h_x^2)h_{yy}}{2(1+h_x^2+h_y^2)^{3/2}}$$