I have come up with something...
Lets think that we have found a solution $u(x,t)$ via the d'Alembert's formula
\begin{equation}
\label{eq:dalamber}
u(x,t) = \frac{u(x-t,0)+u(x+t,0)}{2}+\frac{1}{2}\int\limits_{x-t}^{x+t} u_t (\xi,0) d\xi
\end{equation}
Ofcourse this formula assumes that the functions $u(x-t,0)=\varphi_0(x-t)$ and $u(x+t,0)=\varphi_1(x+t)$ are defined over all the reals. But our given functions $\varphi_0, \varphi_1 : (-\infty, 3) \to \mathbb R$ are not. And so we will have to find suitable continuations of these functions $\hat{\varphi_0}(x),\hat{\varphi_1}(x): \mathbb R \to \mathbb R,$ i.e. functions for which
$$
\hat{\varphi}_0(x)\Big\vert_{(-\infty,3)} = \varphi_0(x), \hspace{1cm} \hat{\varphi}_1(x)\Big\vert_{(-\infty,3)} = \varphi_1(x).
$$
Lets see what the boundary condition $u_x(3,t) =0, t>0$ has to say about the continuations $\hat{\varphi}_0(x),\hat{\varphi}_1(x).$
Differentiating the d'Alembert's formula, and using Leibniz rule for differentiating under the integral sign we get:
\begin{align*}
u_x(x,t) &= \frac{1}{2}\frac{\partial}{\partial x} (\hat{\varphi}_0(x-t)+\hat{\varphi}_0(x+t))+\frac{1}{2} \left[\int\limits_{x-t}^{x+t} \frac{\partial \hat{\varphi}_1(\xi)}{\partial x} d\xi +\hat{\varphi}_1(x+t)-\hat{\varphi}_{1}(x-t) \right] \\
&=\frac{\hat{\varphi}'_0(x-t)+\hat{\varphi}'_0(x+t)}{2}+\frac{\hat{\varphi}_1(x+t)-\hat{\varphi}_1(x-t)}{2}.
\end{align*}
Therefore, since $\xi$ is not dependent of $x$ or $t,$
$$
0 = u_x(3,t) = \frac{\hat{\varphi}'_0(3-t)+\hat{\varphi}'_0(3+t)}{2}+\frac{\hat{\varphi}_1(3+t)-\hat{\varphi}_1(3-t)}{2}.
$$
So in order to $u_x(3,t)=0,t>0$ we need to have
$$\begin{array}{|l}
\frac{\hat{\varphi}'_0(3-t)+\hat{\varphi}'_0(3+t)}{2} =0 \\
\frac{\hat{\varphi}_1(3+t)-\hat{\varphi}_1(3-t)}{2} =0
\end{array}, t >0.$$
More precisely
\begin{align*}
\varphi_1 (x)&=\hat{\varphi}_1 (6-x) \hspace{0.5cm}, x \in (-\infty, 3), \\
\varphi_0(x) &=-\hat{\varphi}'_0 (6-x) \Leftrightarrow \int \varphi'_0(x) dx =\int \hat{\varphi}'_0 (6-x) d(6-x), x \in (-\infty,3), \\
\varphi_0(x) &= \hat{\varphi}'_0(6-x) +C, \hspace{0.5cm} x \in (-\infty,3).
\end{align*}
So suitable continuations are
$$
\hat{\varphi_0}(x) = \begin{cases}
\varphi_0 (x), \hspace{0.5cm} \text{за} \hspace{0.5cm} x \in (-\infty, 3), \\
\varphi_0(6-x), \hspace{0.5cm} \text{за} \hspace{0.5cm} x \in (3,+\infty]
\end{cases}, \hspace{0.5cm}
\hat{\varphi}_1(x) = \begin{cases}
\varphi_1(x), \hspace{0.5cm} x \in (-\infty,3), \\
\varphi_1(6-x), \hspace{0.5cm} x \in (3,+\infty)
\end{cases}.
$$
$$
\hat{u}(x,t) = \frac{\tilde{\varphi}_0(x-t)+\tilde{\varphi}_0(x+t)}{2}+ \frac{1}{2}\int\limits_{x-t}^{x+t} \tilde{\varphi}_1(\xi) d\xi, \hspace{0.5cm} \hat{u}(x,t)\Big\vert_{x<3,t>0} = u(x,t).
$$
