Let $X$ - a random variable with a Poisson distribution with parameter $\lambda >0$. Find certainty equivalent $C[x]$ with respect to the utility function $u(x)=-e^{-x}$.
My try:
$$u(C[X])=\mathbb E(u(X))$$
$$\mathbb E(u(X))=\mathbb E(-e^{-X})=-e^{-\lambda}$$
$$u(C[X])=-e^{-\lambda} \Rightarrow C[X]=u^{-1}(-e^{-\lambda})$$
$$-e^{-x}=y \Rightarrow e^{-x}=-y \Rightarrow -x=\ln (-y) \Rightarrow x =-\ln(-y)$$
$$C[X]=u^{-1}(-e^{-\lambda})=-\ln(-(-e^{-\lambda}))=\lambda$$
Could someone tell me if this reasoning is correct?
The step $E[-e^{-X}] = -e^{-\lambda}$ in second line is false. Most of the time you cannot simply do $E[g(X)] = g(E[X])$ for some arbitrary function $g$; expectations only behave like that for sure if $g$ is linear.
So you have to compute $E[-e^{-X}]$ manually. There is a hack though; if you know the moment generating function of Poisson: $E[e^{tX}] = M_{\text{Poisson}(\lambda)}(t) = e^{\lambda (e^t - 1)}$, you can use that too: $$ E[-e^{-X}] = -E[e^{-X}] = -M_{\text{Poisson}(\lambda)}(-1) = -e^{\lambda(e^{-1} - 1)} $$ The rest of your steps are correct in principal so you should apply them to the above correction to get the right answer.