Can a functional derivative be calculated if we have a function of more than one variable? Can a functional derivative be calculated if we have a function of more than one variable? 
The functional derivative of, for example, $F[b(x)]=e^{\int_0^{x'} dx a(x,y) b(x)}$ is 
\begin{equation}
\frac{\delta F[b(x)]}{\delta b(z)} = a(z,y) e^{\int_0^{x'} dx a(x,y) b(x)}
\end{equation}
But what about if there it's a functional of more than one variable? -  $F[b(x,k)]=e^{\int_0^{x'} \int_0^p  dx dk a(x,y,k) b(x,k) }$? Can we write the following?
\begin{equation}
\frac{\delta F[b(x,k)]}{\delta b(z,k)} = a(z,y,k)e^{\int_0^{x'} \int_0^p  dx dk a(x,y,k) b(x,k )}
\end{equation}
Edit: I just want to note that I've been told that you cannot do this by a lecturer. Therefore, is the answer is yes (or no!), please give some details! (I wasn't convinced that he was correct).
 A: Kostya's response is almost, but not quite correct. It confuses the Gateaux (or Frechet) derivative with the functional derivative. Let F be a functional, acting on a space M of functions defined on a space X. Then the Gateaux derivative of F at the function f is the functional DF[f] such that, to "first order", we have
F[f + $\delta$f] - F[f] = DF[f] [$\delta$f]
To define the functional derivative, we assume that DF[f] can be written as an integral operator:
DF[f][$\delta$f] = $\int$ ${{\delta F} \over {\delta f(x)}} \delta f(x) dx$
In other words, the functional derivative of F at f is the function (not functional!) of x such that:
F[f + $\delta f$] - F[f] = $\int$ ${{\delta F} \over {\delta f(x)}} \delta f(x) dx $
(Since this is a function, I actually prefer to write it as ${{\delta F} \over {\delta f}}(x)$, but, although this makes more sense, it's in violent conflict with the standard notation.)
This definition of the functional derivative always works, as opposed to definitions using multiples of delta functions (which often don't even belong to the set of functions in question!).
Note that since the functional derivative depends the space of functions, not the points of the underlying space, it doesn't make sense to take a partial functional derivative with respect to a particular coordinate direction. In fact, we may be doing this on a manifold, which has no chosen set of coordinates!
However, it is possible to have a functional defined on a product of spaces of functions. For example, with M as above, define F[f,g] = $\int f(x)g(x)^2 dx$. Then it does make sense to define partial functional derivatives with respect to the different spaces of functions. In our example,
F[f, g+$\delta$g] - F[f, g] = $\int f(x)(g(x)+\delta g(x))^2 dx$ - $\int f(x)g(x)^2 dx $
= $\int 2f(x) g(x) \delta g(x) dx $ = $\int {{\partial F} \over {\partial g(x)}} \delta g(x) dx$
That is, at (f, g), the partial functional derivative of F with respect to the second slot is the function:
${{\partial F} \over {\partial g(x)}} = 2f(x)\cdot g(x)$
A: I will not do all the math for you -- just the general idea.
Given a functional $F[g(x_1,x_2,...,x_n)]$. It takes a function of $n$ variables ( I will not write them further) and returns a number. Now you slightly change the argument $F[g+\delta g]$. 
Usually the value of this functional for small $\delta g$ is approximately:
$F[g+\delta g] \simeq F[g]+G[g]\delta g$  
And $G$ is called the functional derivative $G=\frac{\delta F}{\delta g}$ -- compare this to usual deriavative $f(x+\epsilon)\simeq f(x)+\frac{df}{dx}\epsilon$.  So generally you can take a functional derivative over function of any number of variables. An that is the general answer to your general question.

Concerning your example -- there is something strange happening. The expression $\frac{\delta F[b(x,k)]}{\delta b(y,k)}$ just doesn't make any sense to me. Try going step by step in a way I explained. 
A: Jane,
The problem here might be with the exact form of this expression, which seems to have free and bound variables mixed up. Here is an alternative (the problem is with the k).
\begin{equation}
\frac{\delta F[b(x,k)]}{\delta b(z,k')} = a(z,y,k')e^{\int_0^{x'} \int_0^p  dx dk a(x,y,k) b(x,k )}
\end{equation}
