You can find the needed points in each triangle with the calc library: they are the midpoint of each edge and the centroid of the triangles. Then you can make a macro to draw a triangle with those divisions.
Something like this:
\documentclass[tikz,border=2mm]{standalone}
\usetikzlibrary{calc,perspective}
\definecolor{color1}{HTML}{1A5960}
\definecolor{color2}{HTML}{B7DAB9}
\definecolor{color3}{HTML}{AE5689}
\newcommand{\mytriangle}[4] % 3 poiints, 'shadow' opacity
{
\coordinate (center) at ($1/3(#1)+1/3(#2)+1/3*(#3)$);
\coordinate (m12) at ($(#1)!0.5!(#2)$);
\coordinate (m13) at ($(#1)!0.5!(#3)$);
\coordinate (m23) at ($(#2)!0.5!(#3)$);
\draw[fill=color1] (center) -- (m12) -- (#1) -- (m13) -- cycle;
\draw[fill=color2] (center) -- (m12) -- (#2) -- (m23) -- cycle;
\draw[fill=color3] (center) -- (m13) -- (#3) -- (m23) -- cycle;
\draw[thick,fill=black,fill opacity=#4] (#1) -- (#2) -- (#3) -- cycle;
}
\begin{document}
\begin{tikzpicture}[line cap=round,line join=round,3d view={40}{35}]
% Dimensiones
\pgfmathsetmacro\ph{(1+sqrt(5))/2} % golden ratio
\pgfmathsetmacro\ed{3} % edge (half)
\pgfmathsetmacro\hh{\ed*\ph} % height (half)
% Vértices
\coordinate (A1) at ( \hh,-\ed, 0);
\coordinate (B1) at ( \hh, \ed, 0);
\coordinate (C1) at (-\hh, \ed, 0);
\coordinate (D1) at (-\hh,-\ed, 0);
\coordinate (A2) at ( \ed, 0,-\hh);
\coordinate (B2) at (-\ed, 0,-\hh);
\coordinate (C2) at (-\ed, 0, \hh);
\coordinate (D2) at ( \ed, 0, \hh);
\coordinate (A3) at ( 0, \hh,-\ed);
\coordinate (B3) at ( 0, \hh, \ed);
\coordinate (C3) at ( 0,-\hh, \ed);
\coordinate (D3) at ( 0,-\hh,-\ed);
% Faces
\mytriangle{A2}{A1}{D3}{0.4};
\mytriangle{A1}{B1}{A2}{0.5};
\mytriangle{D1}{C3}{D3}{0.2};
\mytriangle{A1}{D3}{C3}{0.3};
\mytriangle{C3}{D2}{A1}{0.0};
\mytriangle{B1}{A1}{D2}{0.1};
\mytriangle{D2}{B3}{B1}{0.3};
\mytriangle{C2}{D1}{C3}{0.2};
\mytriangle{D2}{C3}{C2}{0.1};
\mytriangle{B3}{C2}{D2}{0.2};
\end{tikzpicture}
\end{document}
