If $f$ is some function, then we may define $f_y : x\mapsto f(x+y)$. It is basically a shifted version of the function $f$.
Your equality says that (with $D$ the derivation operator): $ Df(x+y) = Df_{y}(x), $ or, if you prefer $ (Df)_y = D(f_y). $ In other words: "shifting" or "translating" a function commutes with taking derivatives. If we denote by $T_y$ the operator $f\mapsto f_y$, then we could write this as $T_y\circ D = D\circ T_y $
In the language of systems one says that derivation is a time-invariant system. In my opinion it's also a good example of how the $\frac{\partial}{\partial x}$ can sometimes obscure things.
This is always true (if the derived function exists), because (intuitively) derivation only depends on a neighbourhood of the function, and not on the ordinate. In other words, if I give you a picture of a function but forget to draw the $y$-axis, you could still sketch the derived function. (This would be different if I asked you to, for instance, draw $xf(x)$.)
For a more rigorous derivation of this property: let $f$ be any (differentiable) function, then we have that
$ \begin{align*} Df_y(x) &= \lim_{h\to 0}\frac{f_y(x+h)-f_y(h)}{h} \\ &= \lim_{h\to 0} \frac{f(x+h+y)-f(x+y)}{h} \end{align*}$ and $ \begin{align*} (Df)_y(x) &= Df(x+y)\\ &= (Df)(z)|_{z=x+y} \\ &= \lim_{h\to 0}\left.\frac{f(z+h)-f(z)}{h} \right|_{z=x+y} \\ &= \lim_{h\to 0} \frac{f(x+y+h)-f(x+y)}{h} \end{align*} $