Is $\frac{\partial}{\partial x} f(x-y) = - \frac{\partial}{\partial y} f(x-y)$? This seems intuitively plausible to me. But the notation sort of gets in the way when trying to prove this exactly. In particular when using the chain rule to write $\frac{\partial}{\partial y} f(x-y) = - f'(x-y)$ the $'$ looses the information that the chain rule has already been applied.
 A: Let us define $u = x-y$ then we can say
$$
f(x-y) = f(u).
$$
This is great, but we need to transform our derivatives which becomes
$$
\frac{\partial}{\partial x} = \frac{\partial u}{\partial x}\frac{d}{d u} \\
\frac{\partial}{\partial y} = \frac{\partial u}{\partial y}\frac{d}{d u} 
$$
respectively. 
We now sub in the derivatives that we multiple the $u$-derivative by as
$$
\frac{\partial u}{\partial x} = 1\\
\frac{\partial u}{\partial y} = -1
$$
Thus we have
$$
\frac{\partial}{\partial x} f(x-y) = 1\cdot \frac{d}{d u} f(u) = \frac{df}{du}
$$
and
$$
\frac{\partial}{\partial y} f(x-y) = -1\cdot \frac{d}{d u} f(u) = - \frac{df}{du}
$$
or
$$
-\frac{\partial}{\partial y} f(x-y) = \frac{df}{du} = \frac{\partial}{\partial x} f(x-y).
$$
A: I'm not sure that I understand your difficulty, but see if this helps.  It is, I hope, clear that
$$\frac{\partial}{\partial x}\sin(x-y)=\cos(x-y)$$
and
$$\frac{\partial}{\partial y}\sin(x-y)=-\cos(x-y)\ .$$
The case you have described is really no different from this.
Afterthought.  You say that the $f'$ "loses the information that the chain rule has been applied" - well I would say almost the opposite.  It shows clearly that differentiation has been applied.  OK, it isn't clear whether you used the chain rule, the product rule or whatever; but, after all, this is not really important - what is important is that you have calculated the derivative.
A: We can write
$$ g(x,y) = f(z(x,y)) $$
where $z = x-y$. Then $f$ is a single-variable function in $z$ with two parameters $x,y$. Then The notation $f'(x-y)$ means something like
$$ f'(z(x,y)) = \frac{df}{dz} $$
Now if we wanted to find the derivatives $f$ w.r.t $x$ and $y$ instead, then we can employ the multivariable chain rule
$$ \frac{\partial g}{\partial x} = \frac{df}{dz}\frac{\partial z}{\partial x} = 
\frac{df}{dz}\cdot 1 = f'(z) $$
$$ \frac{\partial g}{\partial y} = \frac{df}{dz}\frac{\partial z}{\partial y} = 
\frac{df}{dz}\cdot (-1) = -f'(z) $$
We can revise notation to remove the $z$ dependency and get
$$ \frac{\partial}{\partial x}f(x-y) = f'(x-y) $$
$$ \frac{\partial}{\partial y}f(x-y) = -f'(x-y) $$
In this form, it can be unclear what 'prime' means. It's w.r.t to neither $x$ nor $y$ but instead an implicit variable defined as $z=x-y$. 
If you're used to this short-form notation, you can derive more complicated relationships such as
$$ \frac{\partial}{\partial x}f(x^2+y^2) = 2x\ f'(x^2+y^2) $$
$$ \frac{\partial}{\partial y}f(e^x\sin y) = e^x\cos y \ f'(e^x \sin y) $$
and so on.
