The answer below is very general and doesn't take into account what emerged in later clarifications of the question. It treats $p(x)$, $p(c)$ and $p(x,c)$ as unknown entities. With the clarifications provided, $f$ needs to be considered as a function of many more variables, namely all $p(x,c')$ and all $p(x',c)$. All derivatives should then be partial derivatives, and we can take the derivative of $f$ with respect to $p(x,c)$ while keeping all other $p(x',c)$ and $p(x,c')$ constant. With $p(x)=\sum_cp(x,c)$ and $p(c)=\sum_xp(x,c)$, we have $\partial p(c)/\partial p(x,c)=1$ and $\partial p(x)/\partial p(x,c)=1$, so
$$
\frac{\partial f}{\partial p(x,c)}=\frac1{p(x,c)}+\frac1{p(x)}+\frac1{p(c)}\;.
$$
(You'll need a Lagrange multiplier for the normalization constraint if you take this approach.)
It depends entirely on how you're viewing these probabilities. If you have a model where $p(x)$, $p(c)$ and $p(x,c)$ can be varied separately, you can treat them as distinct independent variables and just take the partial derivative with respect to any of them, e.g. $\partial f/\partial p(x,c)=1/p(x,c)$.
On the other hand, if your model posits some relationship between them, you have to take that relationship into account. If $p(x)$ and $p(c)$ can be expressed as functions of $p(x,c)$, you can form the total derivative with respect to $p(x,c)$, and the result is similar but not quite the same as what you wrote:
$$
\begin{align}
\frac{\mathrm df}{\mathrm dp(x,c)}&=\frac{\partial f}{\partial p(x,c)}+\frac{\partial f}{\partial p(x)}\frac{\mathrm dp(x)}{\mathrm dp(x,c)}+\frac{\partial f}{\partial p(c)}\frac{\mathrm dp(c)}{\mathrm dp(x,c)}
\\
&=\frac1{p(x,c)}+\frac1{p(x)}\frac{\mathrm dp(x)}{\mathrm dp(x,c)}+\frac1{ p(c)}\frac{\mathrm dp(c)}{\mathrm dp(x,c)}\;.
\end{align}
$$
Theoretically you could also consider two of the three as independent and have only one relationship expressing the third in terms of them; in that case, "differentiating with respect to $p(x,c)$" is only well-defined if you specify which of the two others you're holding constant. Then you can form a partial derivative with respect to $p(x,c)$, which you can also obtain with the chain rule. For instance, if you want to differentiate with respect to $p(x,c)$ while holding $p(x)$ constant, that would be
$$
\begin{align}
\left.\frac{\partial f}{\partial p(x,c)}\right|_{p(x)}&=\frac{\partial f}{\partial p(x,c)}\left.\frac{\partial p(x,c)}{\partial p(x,c)}\right|_{p(x)}+\frac{\partial f}{\partial p(x)}\left.\frac{\partial p(x)}{\partial p(x,c)}\right|_{p(x)}+\frac{\partial f}{\partial p(c)}\left.\frac{\partial p(c)}{\partial p(x,c)}\right|_{p(x)}
\\
&=
\frac{\partial f}{\partial p(x,c)}+\frac{\partial f}{\partial p(c)}\left.\frac{\partial p(c)}{\partial p(x,c)}\right|_{p(x)}
\\
&=
\frac1{p(x,c)}+\frac1{p(c)}\left.\frac{\partial p(c)}{\partial p(x,c)}\right|_{p(x)}\;,
\end{align}
$$
where the unqualified partial derivatives are the standard partial derivatives of $f$ with respect to its three arguments while keeping the other two constant.
