From the exercise 8.7 by B. A. Davey, H. A. Priestley Introduction to lattices and order
Let $A_1$, $A_2$, $P$ be CPOs and let $\psi: A_1 \times A_2 \to P $ be a map.
- $\forall a \in A_1$ and $\forall b \in A_2\,\,$ let $\,\,\,\psi^a:A_2 \to P\,\,$ s.t. $\,\,\psi^a(v)= \psi(a,v)\,\,\,$ and $\,\, \psi_b:A_1 \to P\,\,$s.t. $\,\, \psi_b(u) = \psi (u,b)$. $\,\,\,\,$Prove that $\psi$ is continuous (that is iff it preserves the suprema of directed sets) $\iff \psi^a$ and $\psi_b$ are continuous.
Proving this $\Rightarrow$ direction is quite easy, but I am not sure how to conclude the opposite $\Leftarrow$ direction: let $D \in A_1 \times A_2$ be a directed set, then $D = D_1 \times D_2$, with $D_1$ directed in $A_1$ and $D_2$ directed in $A_2$.
If $\bigvee D = (d_1, d_2) = \bigvee(D_1 \times D_2)$, we have: $\psi(\bigvee D) = \psi(d_1,d_2) = \psi^{d_1}(d_2) = \psi_{d_2}(d_1)$, where $$\psi^{d_1}(d_2) = \psi^{\bigvee D_1}(\bigvee D_2) = \bigvee \psi^{\bigvee D_1}(D_2)= \bigvee \psi(\{\bigvee D_1\}\times D_2)$$ and, similarly, $$\psi_{d_2}(d_1) = \psi_{\bigvee D_2}(\bigvee D_1) = \bigvee \psi_{\bigvee D_2}(D_1)= \bigvee \psi(D_1\times \{\bigvee D_2\})$$
how to conclude that $\psi(\bigvee D) = \bigvee \psi(D)$?
I think you can justify the steps (if I didn't get wrong...)
Anyway, I'll start in a point you have already reached.
\begin{align} \psi \left(\bigvee D \right) &= \psi_{d_2}(d_1) = \psi_{d_2} \left( \bigvee D_1 \right)\\ &= \bigvee \psi_{d_2}(D_1) = \bigvee \{ \psi_{d_2}(x) : x \in D_1 \}\\ &= \bigvee \{ \psi(x, d_2) : x \in D_1 \}\\ &= \bigvee \left\{ \psi^{x} \left( \bigvee D_2 \right) : x \in D_1 \right\}\\ &= \bigvee \left\{ \bigvee \left\{ \psi^x(y) : y \in D_2 \right\} : x \in D_1 \right\}\\ &= \bigvee \{ \psi(x,y) : x \in D_1, y \in D_2 \}\\ &= \bigvee \psi(D). \end{align}