How can I build a circuit with this transfer function, using only resistors?
There are two problems with your design.


A voltage divider only produces it's notional output voltage when unloaded.
Your second stage forms an averaging circuit not a sum.


Bottom line is that your approach isn't going to be fruitful.



Fortunately a much simpler design will solve your problem. We just need a weighted average circuit.

For ease of calculation i'm going to assume that when you wrote 0.33 you meant \$\frac{1}{3}\$ if you want to actually use 0.33 that makes the numerics less nice but does'nt change the principles. 

We can rewrite your equation as a weighted average and implement it as a weighted average circuit. This consists of three resistors, one from the input to the output, one from 3.3V to the output and one from ground to the output.

$$V_o = \frac{1}{3}V_i + 1.65 = \frac{2}{6}V_i + \frac{3}{6}3.3+\frac{1}{6}0$$ 

(note that our weights add up to 1 and all of them are positive, that is important)

Now we simply take the reciprocal of the weights to work out our resistor values.

$$R_{Vi} = \frac{6}{2}R$$
$$R_{3.3V} = \frac{6}{3}R$$
$$R_{Ground} = \frac{6}{1}R$$

Where \$R\$ is the output impedance of our weighted average circuit.

Then it becomes a matter of picking a \$R\$ value to set the actual value of our resistors. Ideally we want to pick it such that all three values are standard values. Turns out that \$R=0.5\mathrm{k}\Omega\$ works out nicely giving us.

$$R_{Vi} = \frac{6}{2}R = 1.5\mathrm{k}\Omega$$
$$R_{3.3V} = \frac{6}{3}R = 1\mathrm{k}\Omega$$
$$R_{Ground} = \frac{6}{1}R = 3\mathrm{k}\Omega$$