$\Bbb E(X\mid X^2)=X$ Assume $X$ is exponential distribution. Is the following proof correct?
$$\Bbb E(X\mid X^2)=X$$
This holds because $X$ is measurable in $\sigma\langle  X^2\rangle$$\left(X = \sqrt{X^2}\text{ because } X\ge 0\right)$.

Yes, under the extra condition that $X(\omega)\geq0$ for every $\omega\in\Omega$.
If that condition is not satisfied then - in spite of the fact that $X\geq0$ a.s. - it is possible that $\omega_1,\omega_2\in\Omega$ exist with $0<X(\omega_2)=-X(\omega_1)$.
In that case the set $A:=\{\omega\in\Omega\mid X(\omega)<0\}\in\sigma(X)$ contains $\omega_1$ as element but does not contain $\omega_2$ as element.
This implies that $A$ cannot be an element of $\sigma(X^2)$ and that consequently $X$ is not $\sigma(X^2)$-measurable.
I will prove that by deducing a contradiction on base of the assumption that $A\in\sigma(X^2)$. 
If $A\in\sigma(X^2)$ then  $A=\{\omega\in\Omega\mid X^2(\omega)\in B\}$ for some Borelset $B\subseteq\mathbb R$ which leads to: $$\omega_1\in A\iff X^2(\omega_1)\in B\implies X^2(\omega_2)=X^2(\omega_1)\in B\implies\omega_2\in A$$contradicting that $\omega_2\notin A$.

I would rather go for: $$\mathbb E[X\mid X^2]=|X|$$
This because in the general case we only have $X\geq0$ a.s. and not necessarily $X(\omega)\geq0$ for every $\omega\in\Omega$.
