When implicitly differentiate a function, for example, $f(x)=(G)(x)$, where G is a constant, is it possible to differentiate it such that we can treat G as a variable? From my understanding it is possible, because in the end, after differentiating G with respect to x, you get 0. Technically differentiating a constant with respect to anything other than itself will result in 0, I hope I'm correct on this. If so, it yields:
$f'(x)=(1)(dG/dx)(x)+(G)(1)(dx/dx)$ -> chain rule
because I'm assuming $G$ is a variable. Then...
$f'(x)=G$ because $dG/dx=0$, as $G$ is a constant, and $dx/dx=1$.
In this case, would $dG/dx$ just be 0, because taking the derivative of a constant with respect to anything, other than itself, is 0. In my opinion that would be he only way for $f'(x)=1$.
$G$ is a constant. It's one number, not a variable ranging over numbers. You don't "treat it as a variable", and it's probably misleading or confusing (to yourself) to "assume that it's a varlable". Similarly, you can't "take the derivative of a constant with respect to itself" — it's borderline nonsense to say that or try to think it.
But you can treat $G$ as a function — the constant function $f_G$ whose value is constantly $G$.
The derivative of any constant function is, of course, $0$ – the numerator of the limit that defines the derivative is always $0$.
Fortunately, and as we'd expect, applying the chain rule to $f_G\!\cdot\! x$ gives $f_G$ (here, prime $'$ denotes differentiation with respect to $x$): $$\begin{align} (f_G\!\cdot\! x)' &= f_G\!\cdot\! x' + f_G'\!\cdot\! x \\ &= f_G\!\cdot\! 1 + 0\!\cdot\! x \\ &= f_G. \end{align}$$ More generally, for any function $h$, $(f_G\!\cdot\! h)' = f_G \!\cdot\! h'$, by applying the chain rule just as above (there, $h$ is "the variable $x$", i.e. the identity function, $h(x)=x$ for all $x$). This is completely consistent with $(G\!\cdot\! h)' = G\!\cdot\! h'$.