Questions about updating the secret key of RSA?
if the adversary knows $m^{e_1}$, $m^{e_2}$ and $d_1e_2$, is it able to guess m?


Obviously not; if he could, then they could break RSA.

Here's the RSA problem, given $m^e$, $e$ (and the modulus $N$), recover $m$.

Suppose that we can solve your problem, that is, we had a black box that, given $m^{e_1}$, $m^{e_2}$ and $d_1e_2$, could recover $m$.

Then, given the RSA problem parameters $m^e$ and $e$, we would pick a value $\lambda$ which is relatively prime to $\phi(N)$ (and this can be done simply by picking $\lambda$ prime), and compute $(m^e)^\lambda$.  If we assume $e_1 = e$ and $e_2 = e \cdot \lambda \pmod{\phi(N)}$, we then have $e_2 \cdot d_1 = \lambda \pmod{\phi(N)}$ (as $d_1 = e_1^{-1}$)

We then have $m^{e_1} = m^e$, $m^{e_2} = (m^e)^\lambda$ and $d_1e_2 = \lambda$, we then give these parameters to the black box, and recover $m$, solving the RSA problem.

On the other hand, this argument assumes that (at least) $e_2$ is large (because the above procedure effectively selects a random $e_2$).  If you selected $e_1$, $e_2$ small (say, 16 bits or so), then what the attacker could do is go through the possible $e_1, e_2$ combinations, and recover a billion possible $m$ values, one of which is correct.


  Is the key-update is resistant against leaking old secret keys? If the adversary knows $m^{e_2}, d_1e_2$ and $(e_1,d_1)$, is it able to guess $m$?


No, the key-update is not resistant; the adversary can recover $m$. The values $e_1$, $d_1$ are sufficient to allow him to factor $N$ efficiently, with the knowledge of $d_1e_2$ and $d_1$ and $\phi(N)$, he can recover $e_2$, and with that and $m^{e_2}$ (and again, the factorization of $N$), he can recover $m$.