maximum real power problem
If the objective is to maximize the power dissipated on the load \$R\$. Use, 

\$P_R = \frac{U^2 \cdot R}{(R + 3)^2 + 4^2} = \frac{U^2 \cdot R}{ R^2+6R+9+16} = \frac{U^2}{ R+6+25/R}\$

Then, differentiate the expression to find its maximum, that is, when \$R+6+25/R\$ reaches its minimum.

If you are actually looking for maximizing the power dissipated on both resistors, \$R\$ and \$ 3 \Omega \$, use, 

\$P_{total} = \frac{U^2 \cdot (R+3)}{(R + 3)^2 + 4^2} = \frac{U^2}{\frac{(R + 3)^2 + 16}{R+3}} = \frac{U^2}{R + 3 + \frac{16}{R+3}}\$

Then, find when \$R + 3 + \frac{16}{R+3}\$ is at its smallest value.