Let's see I have the following equation
$$ x=1 $$
I take the derivate of both sides with respect to $x$:
$$ \frac{\partial }{\partial x} x = \frac{\partial }{\partial x}1 $$
Therefore, $1=0$. Clearly, that is not the right approach.
So what is the right way to think of $x=1$. What kind of object is it?
If we're taking derivatives, that means the things on each side of the equals sign are functions. What you've demonstrated is the correct statement that if $f$ is the identity function $f(x) = x$ and $g$ is the constant function $g(x) = 1$, then $f\ne g$.
Partial derivative requires a function as argument therefore if we assume, with little a abuse of notation, $x$ as a function $x= x(x,y,z,\ldots)$ and we states that $x$ is a constant function $x(x,y,z,\ldots)=1$ then
$$\frac{\partial }{\partial x} x(x,y,z,\ldots) = \frac{\partial }{\partial x}1=0$$
Note that of course it should be better to use a different notation for $x$ as a function to distinguish it form the variable $x$, that is for example $\bar x= \bar x(x,y,z,\ldots)$.
You can take derivative of both sides of an identity not an equation.
For example $$\sin^2 x + \cos^2 x =1$$ is an identity, so we can differentiate to get $$2\sin x \cos x -2\sin x\cos x =0$$ or $$\cos 2x = \cos ^2 x - \sin ^2 x $$ gives $$-2\sin 2x = -2\sin x \cos x-2\sin x \cos x $$
Which is $$\sin 2x = 2\sin x \cos x$$
But you can not differentiate the equation $$\sin x =x$$ to get $$\cos x =1$$
Let me throw my hat as well; according to fundamental theorem of calculus, which I assume we all know, derivative is the result of integration,
$$\int_0^1\frac{d}{dx}xdx=\int_0^1dx=[x-x]_0^1=1-0=1$$
I prefer to think integration and derivatives as what they are, area approximations under the curve and slope. I think your approach does not consider that when x is a constant, the derivative is zero, so of course it does not make much sense.
