Let $V$ and $W$ be Brownian motions such that $\mathbb{E}W_tV_t=\rho t$. Let $$R_t=\sup_{u \le t} V_u \mbox{ and } Z_t=\sup_{u \le t}W_u .$$
Show that $$\mathbb{E}R_tZ_t=tf(\rho) .$$
Can you find $f(\rho)$ ?
EDIT:
I have already figured out the first part of the question. Let us fix $t > 0$ and define $\overline{W}_s=\sqrt{t} \cdot W_{\frac{s}{t}}$ and $\overline{V}_s=\sqrt{t} \cdot V_{\frac{s}{t}}$. Let us notice that $\overline{W},\overline{V}$ are both Brownian motions and $\mathbb{E}\overline{W}_s\overline{V}_s=\rho s$. Therefore $\left( W,V \right)\stackrel{d}{=}\left( \overline{W},\overline{V} \right)$.
Hence for $\overline{R}_t=\sup_{u \le t} \overline{V}_u \mbox{ and } \overline{Z}_t=\sup_{u \le t}\overline{W}_u$ we have $\left( R,Z \right)\stackrel{d}{=}\left( \overline{R},\overline{Z} \right)$ and $\overline{R}_t=\sup_{u \le t}\overline{V}_u=\sup_{u \le t} \sqrt{t} \cdot V_{\frac{u}{t}}=\sqrt{t} \cdot R_1$ and analogously $\overline{Z}_t=\sqrt{t} \cdot Z_1$.
Finally $$\mathbb{E}R_tZ_t=\mathbb{E}\overline{R}_t\overline{Z}_t=\mathbb{E} \sqrt{t} \cdot R_1 \cdot \sqrt{t} \cdot Z_1=t\mathbb{E}R_1Z_1$$