RSA: How to calculate the private exponent? I have this problem: In RSA algorithm considering $n=33$ (modulus) and public exponent $e=3$, calculate the corresponding private exponent $d$.

I know that $d = e^{-1} \pmod{\varphi(n)}$ and $\varphi(n) = (p – 1) (q – 1)$ but I don't know $p$ and $q$.

How can I do this?
If I'm understanding your question right, you want to obtain $d$ from given $n$ and $e$.

You'll have to factor $n=33=3*11$ and as $N=p*q$ you have obtained your $p=11$ and $q=3$. Now proceed as usual with calculating the inverse.

As pointed out correctly above, you can't easily generalize this approach to larger numbers as factoring $n$ will be infeasible. Now you see how to factor the number but for larger numbers even the best algorithms fail. Concerning your $e=3$ I'd consider using Coppersmith's attacks on low exponent RSA.