Can partial differential depend on itself? Lets compare $\partial$ and $d$,
for example let $u=e^x$ and $y=e^x$
then $\frac{dy}{dx}=e^x=y$, so we can write that $\frac{d}{dy}\frac{dy}{dx}=1$
but when $\frac{\partial{u}}{\partial{x}}=e^x$ can we write that $\boxed{\frac{\partial{u}}{\partial{x}}=u}$?
would $\frac{\partial}{\partial{u}}\frac{\partial{u}}{\partial{x}}=$ zero or one?
 A: That's exactly why formal language comes into play.
The thing is, it is ok to take the derivatives only with respect to specified arguments of function, so for example, in your case functions $y=y(x)$ and $u=u(x)$ have exactly one arument: $x$.
So
$$
\frac{\partial u(x)}{\partial x} = e^x \quad \text{good} \\
\frac{\partial u(x)}{\partial u} = ?  \quad \text{not good}
$$
I know, in some technical literature sometimes people write something like the 2nd line. But the result heavily depends on what is meant by an expression. Let's take your case as an example.
You'd like to calculate the
$$\frac{\partial}{\partial u} \frac{\partial u}{\partial x}$$
If you mean we should view the $\partial u/\partial x$ as a function of two variables $f(u, x) := \partial u/\partial x=u$ where there is a dependence between arguments $u=u(x)=e^x$, then the derivative $\partial f /\partial u$ is simply $1$, but if you'd like to calculate the derivative $d f/d x$ you'd need to use the chain rule(since the dependence of arguments):
$$
\frac{\partial f}{\partial u} = 1 \\
\frac{d f(u, x)} {d x} = \frac{\partial f(u, x)} {\partial u} \cdot \frac{\partial u} {\partial x} + \frac{\partial f(u, x)} {\partial x} = 1 \cdot e^x + 0 = e^x
$$
But if you had mean that $f(u, x)=\partial u/ \partial x = e^x$  then
$$
\frac{\partial f}{\partial u} = 0 \\
\frac{d f}{d x} = \frac{\partial f}{\partial u} \cdot \frac{d u}{dx} + \frac{\partial f}{\partial x} = 0 \cdot e^x + e^x = e^x
$$
The value of $\frac{\partial }{\partial u} \frac{\partial u}{\partial x}$ is not well defined until you specify how you look at $\frac{\partial u}{\partial x}$ (exactly what I did in the above two cases). The thing is, result depend on that specification.
So, always write a function with its arguments and do not take the derivative w.r.t. variable which does not relate to that function.
