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.