How did we count this open circuit voltage?
Remove \$R_2\$, to give an open circuit in order to determine \$V_{TH}\$.

\$R_3\$ and \$R_4\$ then form a voltage divider across \$U\$, and \$U_2= V_{TH}\$.

Thus, \$V_{TH}=\dfrac{R_3}{R_3 +R_4}U\$. I guess \$U=10 \:V\$.

You can change the resistances into conductances to get the formula in your question, but there's no need to do this. Some prefer to work in G, but mixing the two in one formula is really confusing. Perhaps that was the aim of the question.