Hom functors preserve fibre products. I'm using the definition of fibre product from the Stacks Project.
How do I show that $h_{x \times_y z}(w) = h_{x}(w) \times_{h_y(w)} h_z(w)$ ?
Is equality correct or should it be isomorphism?
I'm guessing isomorphism.  If $p, q$ are the respective maps in the above article, then $-\circ p, -\circ q$ are the corresponding maps in the diagram:
$$
\array {
h_{x}(w) \times_{h_y(w)} h_z(w) & \xrightarrow{-\circ q} & h_z(w) \\
\downarrow - \circ p & \ & \downarrow - \circ g \\
h_{x}(w) & \xrightarrow{-\circ f} & h_y(w)
}
$$
Now we also have maps $-\circ \alpha : h_{x \times_y z} (w) \to h_x(w), \ -\circ \beta : h_{x\times_y z} (w) \to h_z(w)$
Clearly since $f\circ \alpha = g \circ \beta$ in the article, then we have that $-\circ f\circ \alpha = - \circ g \circ \beta$.  
But we need the reverse of this situation namely, given $\beta': h_{x \times_y z}(w)\to h_z(w)$ we need a $\beta$.
But there is an isomorphism $\gamma$ in the article such that $\gamma : w \simeq x \times_y z$.  
Suppose that $-\circ g\circ \beta' = -\circ f\circ \alpha' : h_{x \times_y z}(w) \to h_y(w)$.  Then let $\alpha = p\circ\gamma\circ f \circ \alpha' \circ \gamma^{-1}$.
Then there is a unique morphism $\gamma' : w \to x \times_y z $ such that $p\circ\gamma' = \alpha = p \circ \gamma \circ f \circ \alpha'$. Thus, $\gamma' = \gamma \circ f \circ \alpha'\circ \gamma^{-1}$ and $f\circ \alpha' = \gamma^{-1} \circ \gamma' \circ \gamma : w \to $
 A: As mentioned by @DerekElkins in the comments, this should be an isomorphism. (In fact, it is natural.)
To see this, note that the universal property says that $h_{x\times_y z}(w) = \mathrm{Hom}_\mathcal{C}(w, x \times_y z)$ is in bijective correspondence with the set $$ \{ (\alpha, \beta) \in \mathrm{Hom}_\mathcal{C}(w, x) \times \mathrm{Hom}_\mathcal{C}(w,z) : f \alpha = g \beta \}.$$
On the other hand, this set is precisely the pullback $h_x(w) \times_{h_y(w)} h_z(w)$.
I have gently skimmed some details. In particular, I have mentioned nothing about naturality. Please let me know if you would like me to put them in.
A: $h_{x}(w) \times_{h_y(w)} h_z(w)$ is a pullback of $sets$ and so is the set of pairs $(\alpha,\beta)$ such that the following square commutes: 
\begin{array}{c} h_x(w) \times_{h_y(w)} h_z(w) & \overset{\pi}{\rightarrow} & h_x(w) \\ \ \pi \downarrow&& f^*\downarrow \\ h_z(w) & \overset{g^*}{\rightarrow} & h_y(w) \end{array}
Thus, $f^*\circ \pi\circ \alpha=g^*\circ \pi\circ \beta\Rightarrow f\alpha=g\beta. $
Now, consider the diagram which I copied from your link:

According to the definition of fibre product (pullback), there is an isomorphism 
$\gamma \mapsto(\alpha,\beta)$ such that $f\alpha=g\beta;\ $ i.e.,  $h_{x \times_y z}(w) \cong h_{x}(w) \times_{h_y(w)} h_z(w)$
