Conditional Covariance of Functions of Random Variables $\newcommand{\Cov}{\operatorname{Cov}}
\newcommand{\E}{\mathbb{E}}$
I realize that $\Cov(X,Y) = \E[(X-\mu_X)(Y-\mu_Y)] = \E[\Cov(X,Y|A)] + \Cov(\E[X|A], \E[Y|A])$. But I am not sure how this is applied to functions of random variables conditioned on another random variable belonging to the same probability space.
Let $g(X,Y) = X / (X+Y)$ and $h(X,Y,Z) = (X+Y) / (X+Y+Z)$ , where $X, Y, Z$ are independent Poisson random variables with rates $\lambda_X$, $\lambda_Y$, and $\lambda_Z$, respectively, and where $X + Y + Z > 0$ and $g(X,Y) = 0$ if $Z=N$.
Suppose we know $N = X + Y + Z$. 
What is the conditional covariance of $g(X,Y)$ and $h(X,Y,Z)$ given $N$ and how does one go about finding it?
Possibly this helps (?): For a known $N$, $h(X,Y,Z|N) \sim Binom(N,p) / N$ , where $p$ is the probability of success. This can be shown by recognizing $X+Y$ to be a sum of independent Poisson random variables, and thus also a Poisson random variable with rate $\lambda_X + \lambda_Y$. Conditioned on $(X+Y) + Z = N$, $(X+Y)$ is binomial with $p = (\lambda_X+\lambda_Y)/(\lambda_X + \lambda_Y + \lambda_Z)$.
 A: Based on the HINT from @did, here is a solution for the covariance of $g(X,Y|N)$ and $h(X,Y,Z|N)$.
$\DeclareMathOperator \Cov {Cov}$
$\DeclareMathOperator \E {E}$
$\DeclareMathOperator \Var {Var}$
$$\Cov(g(X,Y), h(X,Y,Z)|N) = \E\biggl[ g(X,Y)h(X,Y,Z)\bigg|N\biggr] - \E\biggl[g(X,Y)\bigg|N\biggr]\ \E\biggl[h(X,Y,Z)\bigg|N\biggr]$$
First consider $\E[X|N]$ $\ldots$
$$\E[X|X+Y+Z] = \E[X|X+W] = \frac{\lambda_X}{\lambda_X + \lambda_W} (X+W) = \frac{\lambda_X}{\lambda_X + \lambda_Y +\lambda_Z} (X+Y+Z) = \frac{\lambda_X}{\lambda_X + \lambda_Y + \lambda_Z} N$$
where we substituted having recognized that the sum of two independent Poisson random variables $Y+Z$ is also a Poisson random variable, say, $W$, whose rate is the sum $\lambda_Y + \lambda_Z$. 
Now, consider the joint expectation 
$$\E\biggl[g(X,Y)h(X,Y,Z)\bigg|N\biggr] = \E\biggl[\frac{X}{X+Y+Z} \mathbf{1}_{X+Y+Z \neq 0} \bigg| X+Y+Z\biggr] \ $$
We can use the property that $\E[aX] = a\E[X]\ $, and hence, 
$$\E\biggl[g(X,Y)h(X,Y,Z)\bigg|N\biggr] = \frac{\lambda_X}{\lambda_X+\lambda_Y+\lambda_Z} \mathbf{1}_{X+Y+Z \neq 0} $$ 
Now, consider $\E[X+Y|N]$ $\ldots$
$$E[X+Y|X+Y+Z] = \E[W|W+Z] = \frac{\lambda_W}{\lambda_W + \lambda_Z} (W+Z) = \frac{\lambda_X+\lambda_Y}{\lambda_X + \lambda_Y + \lambda_Z} N $$
where we made use of another $W$ substitution. Similar to above, 
$$\E \biggl[h(X,Y,Z) \bigg| N \biggr] = \E\biggl[\frac{X+Y}{X+Y+Z} \mathbf{1}_{X+Y+Z \neq 0} \bigg| X+Y+Z \biggr] = \frac{\lambda_X+\lambda_Y}{\lambda_X+\lambda_Y+\lambda_Z} \mathbf{1}_{X+Y+Z \neq 0}$$
Now, one more term remains $\E\biggl[g(X,Y)\bigg| N\biggr]$. We can use the $Law\ of\ Iterated\ Expectations$
$$ \E[X|N] = \E\biggl[\E[X|M]\bigg|N\biggr] $$
where the value of $N$ is determined by $M$. In our case, $N$ is determined by $Z$ and $X+Y$, both existing on the same probability space. So
$$ \E\biggl[g(X,Y) \bigg| N \biggr] = \E \biggl[ \E [g(X,Y)|Z, X+Y)] \bigg| N\biggr] $$
But $g(X,Y)$ is independent of $Z$. In which case, focusing on the inner expectation,
$$ \E \biggl[g(X,Y)\bigg|Z, X+Y)\biggr] = \E \biggl[g(X,Y)\bigg| X+Y) \biggr] = \E \biggl[ \frac{X}{X+Y} \mathbf{1}_{X+Y \neq 0} \bigg| X+Y \biggr] $$
where we have already seen this above.
$$ \E \biggl[ \frac{X}{X+Y} \mathbf{1}_{X+Y \neq 0} \bigg| X+Y \biggr] = \frac{\lambda_X}{\lambda_X + \lambda_Y} \mathbf{1}_{X+Y \neq 0} $$
If we insert this result into the outer expectation, we have
$$ \E \biggl[ g(X,Y) \bigg| N \biggr] = \E \biggl[ \frac{\lambda_X}{\lambda_X + \lambda_Y} \mathbf{1}_{X+Y \neq 0} \bigg| N \biggr] = \frac{\lambda_X}{\lambda_X + \lambda_Y} \mathbf{1}_{X+Y \neq 0}$$
where we used the property that the expectation of a constant is equal to the constant ($\E[b] = b$).
Putting the three terms together, we arrive at our solution for the covariance:
$$ \Cov(g(X,Y), h(X,Y,Z) | N) = \frac{\lambda_X}{\lambda_X + \lambda_Y + \lambda_Z} - \biggl(\frac{\lambda_X}{\lambda_X + \lambda_Y} \cdot \frac{\lambda_X + \lambda_Y}{\lambda_X + \lambda_Y + \lambda_Z} \biggr) = 0$$
So while the functions $g(X,Y)$ and $h(X,Y,Z)$ are clearly dependent, they are not expected to co-vary.
A: Hint: Independent Poisson random variables $U$ and $V$ with respective rates $u$ and $v$ are such that $\mathrm E(U\mid U+V)=\frac{u}{u+v}(U+V)$ hence $\mathrm E\left(\frac{U}{U+V}\,\mathbf 1_{U+V\ne0}\big\vert U+V\right)=\frac{u}{u+v}\,\mathbf 1_{U+V\ne0}$.
