Let P signify that X is algebraic and not equal to 0. Let Q signify that e raised to the X is algebraic. $$1.\exists X(P(X)\land Q(X))$$ $$C.\forall X\exists X_1\left(P\left(X\right)\rightarrow Q\left(X_1\right)\right)\land\forall X\exists X_1\left(Q\left(X_1\right)\rightarrow P\left(X\right)\right)$$ The conclusion is false because $\forall X\exists X_1\left(Q\left(X_1\right)\rightarrow P\left(X\right)\right)$ is false. And it is false because it is equivalent to $\left(\forall X_1Q\left(X_1\right)\right)\rightarrow(\forall X(P\left(X\right))$, which translates to if for all $X_1, e^{X_1}$ is algebraic, then for all X, X is algebraic and not equal to 0. This is plainly false because a counterexample to it is $e^0=1$. Therefore, since the conclusion is false, the premise is false too.
$$1.P\left(X\right)=X\in\mathbb{A}\land X\neq0$$ $$2.Q(X)=e^X\in\mathbb{A}$$ $$3.\exists X(P(X)\land Q(X))\rightarrow\forall X\exists X_1\ (P(X)\leftrightarrow Q(X_1\ ))$$ The consequent of (3) is false because the converse of $\forall X\exists X_1\ (P(X)\rightarrow Q(X_1\ ))$ is false. Therefore, the hypothesis is false.