Proof.
We first show that $(1)$ implies $(2)$. Assume that $U$ is exponentiable, let $V: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$ be an isofibration of simplicial sets, and let $i: A \hookrightarrow B$ be a monomorphism of simplicial sets which is a categorical equivalence; we wish to show that every lifting problem
4.46
\begin{equation} \begin{gathered}\label{equation:exponentiable-equivalent-condition} \xymatrix@R =50pt@C=50pt{ A \ar [d]^{i} \ar [r] & \operatorname{Fun}( \operatorname{\mathcal{C}}/\operatorname{\mathcal{B}}, \operatorname{\mathcal{D}}) \ar [d]^{V \circ } \\ B \ar@ {-->}[ur] \ar [r] & \operatorname{Fun}( \operatorname{\mathcal{C}}/\operatorname{\mathcal{B}}, \operatorname{\mathcal{E}}) } \end{gathered} \end{equation}
admits a solution. Note that the lower horizontal map determines a morphism of simplicial sets $B \rightarrow \operatorname{\mathcal{B}}$. Invoking the universal property of Proposition 4.5.9.5, we can rewrite (4.46) as a lifting problem
4.47
\begin{equation} \begin{gathered}\label{equation:silly-diagram-for-fun2} \xymatrix@R =50pt@C=50pt{ A \times _{\operatorname{\mathcal{B}}} \operatorname{\mathcal{C}}\ar [d]^{j} \ar [r] & \operatorname{\mathcal{D}}\ar [d]^{V} \\ B \times _{\operatorname{\mathcal{B}}} \operatorname{\mathcal{C}}\ar@ {-->}[ur] \ar [r] & \operatorname{\mathcal{E}}. } \end{gathered} \end{equation}
Because $U$ is exponentiable, the left vertical map is a categorical equivalence of simplicial sets. Our assumption that $V$ is an isofibration then guarantees the existence of a solution.
The implication $(2) \Rightarrow (3)$ is immediate. We will complete the proof by showing that $(3)$ implies $(1)$. Assume that condition $(3)$ is satisfied and suppose that we are given a commutative diagram of simplicial sets
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}'' \ar [r]^-{F} \ar [d] & \operatorname{\mathcal{C}}' \ar [r] \ar [d] & \operatorname{\mathcal{C}}\ar [d]^{U} \\ \operatorname{\mathcal{B}}'' \ar [r]^-{ \overline{F} } & \operatorname{\mathcal{B}}' \ar [r] & \operatorname{\mathcal{B}}} \]
where both squares are pullbacks and $\overline{F}$ is a categorical equivalence; we wish to show that $F$ is also a categorical equivalence. By virtue of Exercise 3.1.7.11, there exists a monomorphism of simplicial sets $\iota : \operatorname{\mathcal{B}}'' \hookrightarrow Q$, where $Q$ is a contractible Kan complex. Replacing $\overline{F}$ by the morphism $(\iota , \overline{F}): \operatorname{\mathcal{B}}'' \hookrightarrow Q \times \operatorname{\mathcal{B}}'$ (and $F$ by the morphism $(\iota ,F): \operatorname{\mathcal{C}}'' \hookrightarrow Q \times \operatorname{\mathcal{C}}'$), we can reduce to the case where $\overline{F}$ is a monomorphism of simplicial sets, so that $F$ is also a monomorphism of simplicial sets. To show that $F$ is a categorical equivalence, it will suffice to show that if $V: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{E}}$ is an isofibration of $\infty $-categories, then every lifting problem
4.48
\begin{equation} \begin{gathered}\label{equation:exponentiable-equivalent-condition2} \xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}'' \ar [d]^{F} \ar [r] & \operatorname{\mathcal{D}}\ar [d]^{V} \\ \operatorname{\mathcal{C}}' \ar [r] \ar@ {-->}[ur] & \operatorname{\mathcal{E}}} \end{gathered} \end{equation}
admits a solution (Proposition 4.5.5.4). Invoking the universal property of direct images (Proposition 4.5.9.5), we can rewrite (4.48) as a lifting problem
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{B}}'' \ar [d]^{ \overline{F} } \ar [r] & \operatorname{Fun}( \operatorname{\mathcal{C}}/\operatorname{\mathcal{B}}, \operatorname{\mathcal{D}}) \ar [d]^{V \circ } \\ \operatorname{\mathcal{B}}' \ar [r] \ar@ {-->}[ur] & \operatorname{Fun}( \operatorname{\mathcal{C}}/ \operatorname{\mathcal{B}}, \operatorname{\mathcal{E}}). } \]
Condition $(3)$ guarantees that the right vertical map is an isofibration, so that the solution exists by virtue of our assumption that $\overline{F}$ is a categorical equivalence.
$\square$