We have two functions here: $\begin{align*} &u\colon \mathbb{R}^2\to\mathbb{R}\\ &f\colon\mathbb{R}\to\mathbb{R}. \end{align*}$ The function $u$ is given by $u(x,y) = x-y$, and $z=f\circ u$. So $z$ is a real-valued function of two variables.
But note that $f$ is a real function of real variable. So $f$ does not have partial derivatives, it has regular derivatives, because $f$ is a function of a single variable. The fact that we are evaluating it at a function of two variables does not change the fact that $f$ is a function of a single variable.
Forget for a second about all of these functions. Say you have a function $g\colon\mathbb{R}\to\mathbb{R}$; to fix ideas, say $g(x) = 3x^2-2$. What if $a$ is a constant, and we want to find the derivative of $g(x-a)$ with respect to $x$? We can just use the Chain Rule! Writing $v(x) = x-a$, we have \frac{d}{dx}g(x-a) = \frac{d}{dx}g(v(x)) = \frac{dg}{dv}\frac{dv}{dx} = g'(v(x)))v'(x) = g'(x-a), and since g'(x) = 6x, then \frac{d}{dx}g(v(x)) = g'(x-a) = 6(x-a). Here, $v$ is also a function of a single variable, so derivatives are appropriate.
Now suppose we consider instead $z=g\circ u\colon\mathbb{R}^2\to\mathbb{R}$ (same $u$ as above, $u=x-y$). If we want to find the partial derivative of $z$ with respect to $x$, we do this by fixing $y$; so it's just like trying to take the derivative of $g(x-a)$, except that instead of calling the constant $a$ we call it $y$. So we aren't taking partial derivatives of $g$, we are taking regular derivatives. So $\frac{\partial z}{\partial x} = \frac{d}{dx}g(u(x,y)) = \frac{dg}{du}\frac{\partial u}{\partial x}.$ We use partials for $u$ because $u$ is a function of two variables, but derivatives for $g$ because $g$ is not: it's a function of a single variable.
Back to your example: both $z$ and $u$ are functions of two variables, so we can talk about partials of both $z$ and $u$. But $f$ is a function of a single variable (into which we are plugging the output of $u$), so derivatives are appropriate, not partials.
What is the partial derivative of $z$ with respect to $x$? We consider the function as a function in which $y$ is fixed, and only $x$ changes.