Assume $X,Y$ are stochastic processes satisfying
$$X(t) = X(0) + \int_0^tF_X(s)ds + \int_0^tG_X(s)dW(s) $$ $$Y(t) = Y(0) + \int_0^tF_Y(s)ds + \int_0^tG_Y(s)dW(s) $$
for all $0 \leq t \leq T $. Use the fact that constant processes factor out of integrals, i.e.
$$X(t) = X(0) + tF_x + G_xW(t) \text{ and } Y(t) = Y(0) + tF_Y +G_Y W(t), $$
together with the identities
$$tW(t) = \int_0^t W(s)ds + \int_0^t sdW(s) $$ $$W^2(t) = \int_0^t 1ds + 2\int_0^t W(s)dW(s) $$
to show that $$XY(t)=XY(0)+\int_0^t(X(s)F_Y+Y(s)F_X+G_XG_Y)ds + \int_0^t(X(s)G_Y+Y(s)G_X)dW(s). $$
My question is how should I start this? I thought it would be best to start with $XY$, multiply it out, but I'm not so sure now.
In slightly less verbose notation, you have $$dX_{t}=F_{X}dt+G_{X}dW_{t} \qquad \text{and} \qquad dY_{t}=F_{Y}dt+G_{Y}dW_{t}$$ where I have omitted the argument $t$ on $F_X$ and $G_X$. Ito's product rule yields \begin{align*} d(XY)_{t} & =X_{t}dY_{t}+Y_{t}dX_{t}+d[X,Y]_{t}\\ & =X_{t}\left(F_{Y}dt+G_{Y}dW_{t}\right)+Y_{t}\left(F_{X}dt+G_{X}dW_{t}\right)+G_{X}G_{Y}dt\\ & =\left(X_{t}F_{Y}+Y_{t}F_{X}+G_{X}G_{Y}\right)dt+\left(X_{t}G_{Y}+Y_{t}G_{X}\right)dW_{t} \end{align*} where $[X,Y]$ is the quadratic variation of $X$ and $Y$.