How to calculate \$R_{\text{in}}\$ in this circuit?
You can do a thevenin equivalent of vin, 2R and the first R so that it looks like this: Ux is the voltage divider output of R and 2R and Rx is the parallel combination of 2R and R. 

$$U_x = v_{in}\cdot\frac{R}{R+2R}=v_{in}\cdot\frac{1}{3}$$

And for Rx
$$Rx = \frac{R\cdot 2R}{R+2R} = \frac{2R}{3}$$

So

$$R_{in} = Rx + R = \frac{5R}{3}$$

Finally

$$V_{out} = U_{x}\frac{R1}{R_{in}} = v_{in}\cdot\frac{1}{3}\cdot\frac{3}{5R}\cdot -R1=v_{in} = -\frac{R1}{5R}v_{in}$$

So you can say that $$R_{in} = 5R$$