Why is it contradicting? Let $P$ and $Q$ be functions of $r$ and $r$ be a function of $(x,y,z)$. Also let $f$ be a function of $(x,y)$.
If:
$$P(x,y,z) + f (x,y)= Q(x,y,z) \tag{1} $$
By $(1)$
$$\dfrac{\partial P}{\partial x}  \neq \dfrac{\partial Q}{\partial x} \Rightarrow \dfrac{dP}{dr} \dfrac{\partial r}{\partial x} \neq
\dfrac{dQ}{dr} \dfrac{\partial r}{\partial x}\Rightarrow \dfrac{dP}{dr} \neq \dfrac{dQ}{dr}  \tag{2} $$
Also by $(1)$
$$\dfrac{\partial P}{\partial z}  = \dfrac{\partial Q}{\partial z}\Rightarrow \dfrac{dP}{dr} \dfrac{\partial r}{\partial z} =
\dfrac{dQ}{dr} \dfrac{\partial r}{\partial z} \Rightarrow \dfrac{dP}{dr} =
\dfrac{dQ}{dr}  \tag{3}$$
$(2)$ and $(3)$ contradict. Why is this so?
 A: You say that: $\qquad P(r(x,y,z))+f(x,y)=Q(r(x,y,z))$
Which is to say: $\quad f(x,y)=[Q-P]\circ r(x,y,z)$
Therefore  $\qquad\quad~~ 0~{= \dfrac{\partial f(x,y)}{\partial z}\\=\left[\dfrac{\mathsf d [Q-P](r)}{\mathsf dr\qquad\qquad}\right]\!\!(x,y,z)\cdot\dfrac{\partial r(x,y,z)}{\partial z\qquad\quad}}$
So either $\dfrac{\mathsf d [Q-P](r)}{\mathsf dr\qquad\qquad}=0$ or $\dfrac{\partial r(x,y,z)}{\partial z\qquad\quad} =0$
In the first case, $f(x,y)$ must be a constant, and in the second $r$ is invariant with respect to $z$.   In either case, there is no contradiction involved. 
If $f(x,y)$ is constant, then $\dfrac{\partial P}{\partial x}+0=\dfrac{\partial Q}{\partial x}$.
If $\dfrac{\partial r(x,y,z)}{\partial z\qquad\quad}=0$ then $\dfrac{\partial P}{\partial z}=\dfrac{\partial Q}{\partial z}=0$.
A: If 
$$P(r(x,y,z)) + f(x,y)= Q(r(x,y,z)) \tag{1}$$ 
then
$$\dfrac{d P(r)}{d r}\dfrac{\partial r}{\partial x}+\dfrac{\partial f}{\partial x}=\dfrac{d Q(r)}{d r}\dfrac{\partial r}{\partial x} \tag{2}$$
and
$$\dfrac{d P(r)}{d r}\dfrac{\partial r}{\partial z}=\dfrac{d Q(r)}{d r}\dfrac{\partial r}{\partial z} \tag{3}.$$
In my opinion (2) does not  contradict (3):  if $\dfrac{\partial r}{\partial z}\not=0$ then, from (3), $\dfrac{d P(r)}{d r}=\dfrac{d Q(r)}{d r}$ and, by (2), it follows that $\dfrac{\partial f}{\partial x}=0$. In the same way $\dfrac{\partial f}{\partial y}=0$. 
The difference $P(r)-Q(r)=f$ could be identically constant. 
